39 条题解

  • 5
    @ 2016-12-21 19:24:06

    因为它的状态比较多,比较吓人。(但是这题有错误,夫妻条件不管才能过。)

    说说正解吧。

    dp[i][j][h][k]表示前i个房间,住了j对夫妻,一共住了h个男的,k个女的这个状态的最小费用。

    然后就是暴力枚举每一个房间,更新数组了。

    这里有一个很有趣的地方就是,夫妻房最多用1对就已经最优了,因为你用两对的话,可以把他们拆散,2个男的一件,2个女的一件。

    所以这里如果考虑夫妻的,是可以卡到你超时的,如果注意到夫妻 只能用一对,那就不会超时。

    • @ 2016-12-21 19:24:31

      #include <cstdio>
      #include <cstdlib>
      #include <cstring>
      #include <cmath>
      #include <algorithm>
      #include <assert.h>
      #define IOS ios::sync_with_stdio(false)
      using namespace std;
      #define inf (0x3f3f3f3f)
      typedef long long int LL;

      #include <iostream>
      #include <sstream>
      #include <vector>
      #include <set>
      #include <map>
      #include <queue>
      #include <string>
      int dp[300 + 2][300 + 2][300 + 2];
      const int maxn = 300 + 20;
      int a[maxn];
      int cost[maxn];
      void work() {
      int nan, nv, room, toget;
      cin >> nan >> nv >> room >> toget;
      for (int i = 1; i <= room; ++i) {
      cin >> a[i] >> cost[i];
      }
      memset(dp, 0x3f, sizeof dp);
      dp[0][0][0] = 0;
      toget = min(toget, 1);
      toget = 0;
      for (int i = 1; i <= room; ++i) {
      for (int j = toget; j >= 0; --j) {
      for (int h = nan; h >= 0; --h) {
      for (int k = nv; k >= 0; --k) {
      if (k >= a[i]) {
      dp[j][h][k] = min(dp[j][h][k], dp[j][h][k - a[i]] + cost[i]);
      } else {
      dp[j][h][k] = min(dp[j][h][k], dp[j][h][0] + cost[i]);
      }
      if (h >= a[i]) {
      dp[j][h][k] = min(dp[j][h][k], dp[j][h - a[i]][k] + cost[i]);
      } else {
      dp[j][h][k] = min(dp[j][h][k], dp[j][0][k] + cost[i]);
      }
      if (j >= 1 && h >= 1 && k >= 1) {
      dp[j][h][k] = min(dp[j][h][k], dp[j - 1][h - 1][k - 1] + cost[i]);
      }
      // dp[0][h][k] = min()
      }
      }
      }
      }
      // cout << dp[0][2][1] << endl;
      int ans = inf;
      for (int i = 0; i <= toget; ++i) {
      ans = min(ans, dp[i][nan][nv]);
      }
      if (ans == inf) {
      cout << "Impossible" << endl;
      return;
      }
      cout << ans << endl;
      }

      int main() {
      #ifdef local
      freopen("data.txt", "r", stdin);
      // freopen("data.txt", "w", stdout);
      #endif
      work();
      return 0;
      }

  • 1
    @ 2020-05-15 22:46:50

    背包动规

    #include<iostream>
    #include<cstring>
    using namespace std;
    int b[301],p[301];
    int dp[601][601][2];
    int main()
    {
        int m,f,r,c;
        cin>>m>>f>>r>>c;
        memset(dp,127,sizeof(dp));
        for(int i=1;i<=r;i++)
           cin>>b[i]>>p[i];
        dp[0][0][0]=0;
        for(int l=1;l<=r;l++)
               for(int i=m;i>=0;i--) 
                   for(int j=f;j>=0;j--)
                   {
                       if(b[l]>=2)dp[i][j][1]=min(dp[i][j][0]+p[l],dp[i][j][1]);
                       for(int o=1;o<=b[l];o++)
                       {
                           if(i+o<=m+c&&i-o>=0)
                           dp[i][j][0]=min(dp[i][j][0],dp[i-o][j][0]+p[l]);
                           if(j+o<=f+c&&j-o>=0)
                           dp[i][j][0]=min(dp[i][j][0],dp[i][j-o][0]+p[l]);
                       }
                   }
        int ans=9999999;
            ans=min(dp[m-1][f-1][1],dp[m][f][0]);
        if(ans!=9999999)cout<<ans;
        else cout<<"Impossible";
        return 0;
    }
    
  • 1
    @ 2018-02-05 20:50:04

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <assert.h>
    #define IOS ios::sync_with_stdio(false)
    using namespace std;
    #define inf (0x3f3f3f3f)
    typedef long long int LL;
    #include <iostream>
    #include <sstream>
    #include <vector>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    int dp[300 + 2][300 + 2][300 + 2];
    const int maxn = 300 + 20;
    int a[maxn];
    int cost[maxn];
    void work() {
    int nan, nv, room, toget;
    cin >> nan >> nv >> room >> toget;
    for (int i = 1; i <= room; ++i) {
    cin >> a[i] >> cost[i];
    }
    memset(dp, 0x3f, sizeof dp);
    dp[0][0][0] = 0;
    toget = min(toget, 1);
    toget = 0;
    for (int i = 1; i <= room; ++i) {
    for (int j = toget; j >= 0; --j) {
    for (int h = nan; h >= 0; --h) {
    for (int k = nv; k >= 0; --k) {
    if (k >= a[i]) {
    dp[j][h][k] = min(dp[j][h][k], dp[j][h][k - a[i]] + cost[i]);
    } else {
    dp[j][h][k] = min(dp[j][h][k], dp[j][h][0] + cost[i]);
    }
    if (h >= a[i]) {
    dp[j][h][k] = min(dp[j][h][k], dp[j][h - a[i]][k] + cost[i]);
    } else {
    dp[j][h][k] = min(dp[j][h][k], dp[j][0][k] + cost[i]);
    }
    if (j >= 1 && h >= 1 && k >= 1) {
    dp[j][h][k] = min(dp[j][h][k], dp[j - 1][h - 1][k - 1] + cost[i]);
    }
    // dp[0][h][k] = min()
    }
    }
    }
    }
    // cout << dp[0][2][1] << endl;
    int ans = inf;
    for (int i = 0; i <= toget; ++i) {
    ans = min(ans, dp[i][nan][nv]);
    }
    if (ans == inf) {
    cout << "Impossible" << endl;
    return;
    }
    cout << ans << endl;
    }
    int main() {
    #ifdef local
    freopen("data.txt", "r", stdin);
    // freopen("data.txt", "w", stdout);
    #endif
    work();
    return 0;
    }

  • 1
    @ 2017-08-16 22:02:13

    大家都写的这种版本
    #include<bits/stdc++.h>
    using namespace std;
    #define xyf main
    int m,f,r,c;
    int b[301],p[301];
    int dp[301][301][2];
    int xyf()
    {
    cin>>m>>f>>r>>c;
    memset(dp,0x3f3f3f,sizeof(dp));
    for(int i=1;i<=r;++i)
    cin>>b[i]>>p[i];
    dp[0][0][0]=0;
    for(int l=1;l<=r;++l)
    for(int i=m;i>=0;--i)
    for(int j=f;j>=0;--j)
    {
    if(b[l]>=2)
    dp[i][j][1]=min(dp[i][j][0]+p[l],dp[i][j][1]);
    for(int o=1;o<=b[l];++o)
    {
    if(o+i<=m+c&&o<=i)
    dp[i][j][0]=min(dp[i][j][0],dp[i-o][j][0]+p[l]);
    if(o+j<=f+c&&o<=j)
    dp[i][j][0]=min(dp[i][j][0],dp[i][j-o][0]+p[l]);
    }
    }
    int ans=0x3f3f3f;
    ans=min(dp[m-1][f-1][1],dp[m][f][0]);
    if(ans!=0x3f3f3f)
    cout<<ans<<endl;
    else
    cout<<"Impossible\n";
    return 0;
    }
    我来个滚动数组,快3倍。
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    using namespace std;
    typedef long long LL;
    const int N = 300 + 24;
    int m,f,r,c;
    int dp[N][N][2];
    int main()
    {
    scanf("%d%d%d%d",&m,&f,&r,&c);

    memset(dp,0x3f3f3f3f,sizeof(dp));
    dp[m][f][1] = dp[m][f][0] = 0;
    for(int i = 0; i < r; i++)
    {
    int a,b;
    scanf("%d%d",&a,&b);
    for(int i = 0; i <= m; i++)
    {
    for(int j = 0; j <= f; j++)
    {
    int x1 = max(0,i-a);
    int x2 = max(0,j-b);
    dp[x1][j][0] = min(dp[x1][j][0],dp[i][j][0] + b);
    dp[i][x2][0] = min(dp[i][x2][0],dp[i][j][0] + b);
    dp[x1][j][1] = min(dp[x1][j][1],dp[i][j][1] + b);
    dp[i][x2][1] = min(dp[i][x2][1],dp[i][j][1] + b);
    if(i > 0 && j > 0 && a >= 2)
    dp[i-1][j-1][1] = min(dp[i-1][j-1][1],dp[i][j][0] + b);
    }
    }
    }
    int d = min(dp[0][0][0],dp[0][0][1]);
    if(d == 0x3f3f3f3f)
    printf("Impossible");
    else
    printf("%d",d);
    }

  • 1
    @ 2016-11-15 20:39:13
    program vijosP1240;
    
    var
         k1,k2,n,c:longint;
         b,p:array[1..300] of longint;
         f:array[0..300,0..300,0..300,0..1] of longint;
         ans:longint;
    
    function min(a,b:longint):longint;
    begin
         if a<b then
            exit(a)
         else
            exit(b);
    end;
    
    function max(a,b:longint):longint;
    begin
         if a>b then
            exit(a)
         else
            exit(b);
    end;
    
    procedure init;
    var
       i:longint;
    begin
         readln(k1,k2,n,c);
         for i:=1 to n do
             readln(b[i],p[i]);
    end;
    
    procedure dp;
    var
       i,t1,t2:longint;
    begin
         ans:=maxlongint;
         fillchar(f,sizeof(f),127);
         for i:=0 to n do
             f[i,0,0,0]:=0;
         for i:=1 to n do
             for t1:=0 to k1 do
                 for t2:=0 to k2 do
                 begin
                      if (t1=0) and (t2=0) then continue;
                      f[i,t1,t2,0]:=min(f[i-1,max(0,t1-b[i]),t2,0]+p[i]
                                       ,f[i-1,t1,max(0,t2-b[i]),0]+p[i]);
                      f[i,t1,t2,0]:=min(f[i,t1,t2,0]
                                        ,f[i-1,t1,t2,0]);
                      //if c>=1 then
                      //begin
                      //f[i,t1,t2,1]:=min(f[i-1,max(0,t1-b[i]),t2,1]+p[i]
                      //                ,f[i-1,t1,max(0,t2-b[i]),1]+p[i]);
                      //f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1,t2,1]);
                      //if (t1>0) and (t2>0) then
                      //f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1-1,t2-1,0]+p[i]);
                      //end;
                 end;
         for i:=1 to n do
             //for t1:=0 to 1 do
             //    if (t1=0) or ((t1=1) and (c>=1)) then
                    ans:=min(ans,f[i,k1,k2,0]);
         if ans>=21*1000*1000 then
            writeln('impossible')
         else
            writeln(ans);
    end;
    
    begin
         init;
         dp;
    end.
    
  • 1
    @ 2015-11-01 20:02:17

    真是太诡异了我g[I][j][k] 打成了g[I][k][k]对了90分,改成g[I][j][k]居然0分!

  • 0
    @ 2016-11-15 20:38:36

    ···pascal
    program vijosP1240;

    var
    k1,k2,n,c:longint;
    b,p:array[1..300] of longint;
    f:array[0..300,0..300,0..300,0..1] of longint;
    ans:longint;

    function min(a,b:longint):longint;
    begin
    if a<b then
    exit(a)
    else
    exit(b);
    end;

    function max(a,b:longint):longint;
    begin
    if a>b then
    exit(a)
    else
    exit(b);
    end;

    procedure init;
    var
    i:longint;
    begin
    readln(k1,k2,n,c);
    for i:=1 to n do
    readln(b[i],p[i]);
    end;

    procedure dp;
    var
    i,t1,t2:longint;
    begin
    ans:=maxlongint;
    fillchar(f,sizeof(f),127);
    for i:=0 to n do
    f[i,0,0,0]:=0;
    for i:=1 to n do
    for t1:=0 to k1 do
    for t2:=0 to k2 do
    begin
    if (t1=0) and (t2=0) then continue;
    f[i,t1,t2,0]:=min(f[i-1,max(0,t1-b[i]),t2,0]+p[i]
    ,f[i-1,t1,max(0,t2-b[i]),0]+p[i]);
    f[i,t1,t2,0]:=min(f[i,t1,t2,0]
    ,f[i-1,t1,t2,0]);
    //if c>=1 then
    //begin
    //f[i,t1,t2,1]:=min(f[i-1,max(0,t1-b[i]),t2,1]+p[i]
    // ,f[i-1,t1,max(0,t2-b[i]),1]+p[i]);
    //f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1,t2,1]);
    //if (t1>0) and (t2>0) then
    //f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1-1,t2-1,0]+p[i]);
    //end;
    end;
    for i:=1 to n do
    //for t1:=0 to 1 do
    // if (t1=0) or ((t1=1) and (c>=1)) then
    ans:=min(ans,f[i,k1,k2,0]);
    if ans>=21*1000*1000 then
    writeln('impossible')
    else
    writeln(ans);
    end;

    begin
    init;
    dp;
    end.
    ```

  • 0
    @ 2016-11-15 20:35:54

    AC100留念,顺便说下这题数据有问题,考虑了夫妻反而WA....下面程序注释掉就是考虑夫妻的情况
    '''pascal
    program vijosP1240;

    var
    k1,k2,n,c:longint;
    b,p:array[1..300] of longint;
    f:array[0..300,0..300,0..300,0..1] of longint;
    ans:longint;

    function min(a,b:longint):longint;
    begin
    if a<b then
    exit(a)
    else
    exit(b);
    end;

    function max(a,b:longint):longint;
    begin
    if a>b then
    exit(a)
    else
    exit(b);
    end;

    procedure init;
    var
    i:longint;
    begin
    readln(k1,k2,n,c);
    for i:=1 to n do
    readln(b[i],p[i]);
    end;

    procedure dp;
    var
    i,t1,t2:longint;
    begin
    ans:=maxlongint;
    fillchar(f,sizeof(f),127);
    for i:=0 to n do
    f[i,0,0,0]:=0;
    for i:=1 to n do
    for t1:=0 to k1 do
    for t2:=0 to k2 do
    begin
    if (t1=0) and (t2=0) then continue;
    f[i,t1,t2,0]:=min(f[i-1,max(0,t1-b[i]),t2,0]+p[i]
    ,f[i-1,t1,max(0,t2-b[i]),0]+p[i]);
    f[i,t1,t2,0]:=min(f[i,t1,t2,0]
    ,f[i-1,t1,t2,0]);
    if c>=1 then
    begin
    f[i,t1,t2,1]:=min(f[i-1,max(0,t1-b[i]),t2,1]+p[i]
    ,f[i-1,t1,max(0,t2-b[i]),1]+p[i]);
    f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1,t2,1]);
    if (t1>0) and (t2>0) then
    f[i,t1,t2,1]:=min(f[i,t1,t2,1],f[i-1,t1-1,t2-1,0]+p[i]);
    end;
    end;
    for i:=1 to n do
    for t1:=0 to 1 do
    if (t1=0) or ((t1=1) and (c>=1)) then
    ans:=min(ans,f[i,k1,k2,t1]);
    if ans>=21*1000*1000 then
    writeln('impossible')
    else
    writeln(ans);
    end;

    begin
    init;
    dp;
    end.
    '''

  • 0
    @ 2016-10-03 15:53:50

    果然水题需细心。。
    ```c++
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;

    int m, f, r, c, dp[305][305][305];
    int b[305], p[305];
    int main()
    {
    scanf("%d%d%d%d", &m, &f, &r, &c);
    memset(dp, 127/3, sizeof dp);
    for (int i = 1; i <= r; i++)
    scanf("%d%d", &b[i], &p[i]);
    dp[0][0][0] = 0; // 前i间j男k女
    for (int i = 1; i <= r; i++)
    for (int j = 0; j <= m; j++)
    for (int k = 0; k <= f; k++) {
    dp[i][j][k] = min(dp[i-1][j][k],
    min(dp[i-1][max(j-b[i], 0)][k], dp[i-1][j][max(k-b[i], 0)])+p[i]);
    if (j == 1 && k == 1 && c >= 1 && b[i] >= 2)
    dp[i][j][k] = min(dp[i][j][k], dp[i-1][0][0]+p[i]);
    }
    if (dp[r][m][f] <= 10000000)
    cout << dp[r][m][f] << endl;
    else
    cout << "Impossible" << endl;
    return 0;
    }

  • 0
    @ 2016-05-24 13:31:19

    数据弱了吧?

  • 0
    @ 2015-12-09 20:13:24

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    using namespace std;
    typedef long long LL;
    const int N = 300 + 24;
    int m,f,r,c;
    int dp[N][N][2];
    int main()
    {
    #ifdef CDZSC_OFFLINE
    freopen("in.txt","r",stdin);
    #endif //CDZSC_OFFLINE
    while(~scanf("%d%d%d%d",&m,&f,&r,&c))
    {
    memset(dp,0x3f3f3f3f,sizeof(dp));
    dp[m][f][1] = dp[m][f][0] = 0;
    for(int i = 0; i < r; i++)
    {
    int a,b;
    scanf("%d%d",&a,&b);
    for(int i = 0; i <= m; i++)
    {
    for(int j = 0; j <= f; j++)
    {
    int x1 = max(0,i-a);
    int x2 = max(0,j-a);
    dp[x1][j][0] = min(dp[x1][j][0],dp[i][j][0] + b);
    dp[i][x2][0] = min(dp[i][x2][0],dp[i][j][0] + b);
    dp[x1][j][1] = min(dp[x1][j][1],dp[i][j][1] + b);
    dp[i][x2][1] = min(dp[i][x2][1],dp[i][j][1] + b);
    if(i > 0 && j > 0 && a >= 2)
    dp[i-1][j-1][1] = min(dp[i-1][j-1][1],dp[i][j][0] + b);
    }
    }
    }
    int d = min(dp[0][0][0],dp[0][0][1]);
    if(d == 0x3f3f3f3f)
    printf("Impossible\n");
    else
    printf("%d\n",d);
    }
    }

  • 0
    @ 2014-10-11 21:31:25

    /*【题解】
    动态规划;
    设f[i][j][k][0]表示前i个房间,男的放了j个女的放了k个的最优值,且没有夫妻在同一房间中
    设f[i][j][k][1]表示前i个房间,男的。。。。。。。。。。。。。,且有一对夫妻在同一房间中。
    为什么状态只有一对呢?
    这是因为,如果有两对夫妻分别在不同的房间,那么完全可以将它们拆散,使得两男两女分别在两个不同的房间。这样可以使房间的
    容量得到最大限度的利用。
    转移方程
    f[i][j][k][0]=min(f[i-1][j][k][0],f[i-1][j-v][k][0]+pi,f[i-1][j][k-v][0]+pi);
    f[i][j][k][1]=min(f[i-1][j][k][1],f[i-1][j-1][k-1][0]+pi,f[i-1][j-v][k][1]+pi,f[i-1][j][k-v][1]+pi);
    注意不要漏掉状态,否则前功尽弃!
    至于网上流传的一个三维的数组,请试试以下数据。
    【*test.in】
    6 6 5 2
    2 5
    2 6
    3 5
    5 1
    5 2
    【*test.out】
    8
    这是我在测试网上流传的题解程序时发现的问题,程序会出现10以上的解。
    */

    #include <iostream>
    #include <cstring>
    using namespace std;
    int m,f,r,c,b[301],p[301],ff[301][301][301][2];
    int main()
    {
    cin>>m>>f>>r>>c;
    for (int i=1;i<=r;i++)
    cin>>b[i]>>p[i];
    memset(ff,127,sizeof(ff));
    int temp=ff[0][0][1][1],ans=temp;
    ff[0][0][0][0]=0;
    for (int i=1;i<=r;i++)
    for (int j=m;j>=0;j--)
    for (int k=f;k>=0;k--)
    {
    if (b[i]>=2 && j>=1 && k>=1)
    ff[i][j][k][1]=min(ff[i-1][j-1][k-1][0]+p[i],ff[i-1][j][k][1]);
    ff[i][j][k][0]=min(ff[i][j][k][0],ff[i-1][j][k][0]);
    for (int t=1;t<=b[i];t++)
    {
    if (j-t>=0)
    {
    ff[i][j][k][0]=min(ff[i][j][k][0],ff[i-1][j-t][k][0]+p[i]);
    ff[i][j][k][1]=min(ff[i][j][k][1],ff[i-1][j-t][k][1]+p[i]);
    }
    if (k-t>=0)
    {
    ff[i][j][k][0]=min(ff[i][j][k][0],ff[i-1][j][k-t][0]+p[i]);
    ff[i][j][k][1]=min(ff[i][j][k][1],ff[i-1][j][k-t][1]+p[i]);
    }
    }
    }
    ans=min(ff[r][m][f][0],ff[r][m][f][1]);
    if (ans==temp) cout <<"Impossible"<<endl;else cout<<ans<<endl;
    return 0;
    }

  • 0
    @ 2014-03-29 21:29:57

    夫妻要同住的话只要一对夫妻就可以了,因为2对夫妻的话只要2间房一个住2男 一个住2女就可以了^_^

  • 0
    @ 2013-07-14 17:09:58

    测试数据 #0: Accepted, time = 0 ms, mem = 900 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 896 KiB, score = 10

    测试数据 #2: Accepted, time = 0 ms, mem = 896 KiB, score = 10

    测试数据 #3: Accepted, time = 171 ms, mem = 892 KiB, score = 10

    测试数据 #4: Accepted, time = 187 ms, mem = 896 KiB, score = 10

    测试数据 #5: Accepted, time = 171 ms, mem = 896 KiB, score = 10

    测试数据 #6: Accepted, time = 171 ms, mem = 892 KiB, score = 10

    测试数据 #7: Accepted, time = 171 ms, mem = 892 KiB, score = 10

    测试数据 #8: Accepted, time = 156 ms, mem = 896 KiB, score = 10

    测试数据 #9: Accepted, time = 171 ms, mem = 892 KiB, score = 10

  • 0
    @ 2012-09-18 20:10:40

    orz楼上神牛

  • 0
    @ 2012-07-20 21:27:05

    又被题目坑了

    这题根本不需要考虑夫妻同房,考虑了就WA!

    强烈鄙视出题人

  • 0
    @ 2009-11-04 09:25:48

    方程写错了,结果全过了。。。

  • 0
    @ 2009-11-04 08:21:11

    正确做法要考虑夫妻。

    易知最多只能有一对夫妇一个房间

    f表示前i个房间住j名男性k名女性并且没有夫妇住在一起的最小花费

    f表示前i个房间住j名男性k名女性并且有一对夫妇住在一起的最小花费

    f=min(f,f+p[i],f+p[i])

    f=min(f,f+p[i],f[i-1,j,k-v[i],1+p[i],f+p[i])

    约束条件很多要小心。

    http://blog.sina.com.cn/s/blog_5774b8650100fv1b.html

  • 0
    @ 2009-10-29 11:59:34

    没有考虑夫妻,直接二维背包

  • 0
    @ 2009-08-24 20:13:47

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 994ms

    ├ 测试数据 05:答案正确... 994ms

    ├ 测试数据 06:答案正确... 994ms

    ├ 测试数据 07:答案正确... 978ms

    ├ 测试数据 08:答案正确... 994ms

    ├ 测试数据 09:答案正确... 1025ms

    ├ 测试数据 10:答案正确... 1009ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:6988ms

    万慢

信息

ID
1240
难度
5
分类
动态规划 | 背包 点击显示
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递交数
676
已通过
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通过率
36%
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