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301 条题解

  • 0
    @ 2015-08-21 10:09:40

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int main()
    {
    int n,v,w[101],p[101],f[101][1001];
    scanf("%d%d",&n,&v);
    for(int i=1;i<=n;i++)
    scanf("%d%d",&p[i],&w[i]);
    int ans=-1;
    f[0][0]=0;
    for(int i=1;i<=v;i++)
    f[0][i]=-1;
    for(int i=1;i<=n;i++)
    for(int j=0;j<=v;j++)
    {
    if(j-w[i]>=0)
    {
    int pd1=f[i-1][j];
    int pd2=f[i-1][j-w[i]];
    if(pd1!=-1&&pd2!=-1)
    f[i][j]=max(pd1,pd2+p[i]);
    else
    if(pd1==-1&&pd2!=-1)
    f[i][j]=pd2+p[i];
    else
    if(pd1!=-1&&pd2==-1)
    f[i][j]=pd1;
    else
    f[i][j]=-1;
    }
    else
    {
    if(f[i-1][j]!=-1)
    f[i][j]=f[i-1][j];
    else
    f[i][j]=-1;
    }
    if(i==n)
    {
    if(f[i][j]>ans)
    ans=f[i][j];
    }
    }
    printf("%d\n",ans);
    return 0;
    }

  • 0
    @ 2015-08-20 20:08:58

    裸的背包问题
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    int main() {
    int T,M,time[101],value[101];
    int solution[1001][101];
    scanf("%d%d",&T,&M);
    for(int i=1;i<=M;i++) scanf("%d%d",&time[i],&value[i]);
    memset(solution,0,sizeof(solution));
    for(int i=1;i<=T;i++)
    for(int j=1;j<=M;j++) {
    if(time[j]>i) solution[i][j]=solution[i][j-1];
    else solution[i][j]=std::max(solution[i][j-1],value[j]+solution[i-time[j]][j-1]);
    }
    printf("%d",solution[T][M]);
    return 0;
    }

  • 0
    @ 2015-08-20 15:42:52

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #2: Accepted, time = 0 ms, mem = 5224 KiB, score = 10

    测试数据 #3: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #4: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #5: Accepted, time = 0 ms, mem = 5212 KiB, score = 10

    测试数据 #6: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #7: Accepted, time = 0 ms, mem = 5220 KiB, score = 10

    测试数据 #8: Accepted, time = 0 ms, mem = 5216 KiB, score = 10

    测试数据 #9: Accepted, time = 0 ms, mem = 5220 KiB, score = 10

    Accepted, time = 0 ms, mem = 5224 KiB, score = 100

    简单的背包问题

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    using namespace std;
    typedef long long LL;
    const int N = 1100;
    const int M = 150;
    LL dp[N];
    int pa[N][N];
    int n,m,t;
    LL num;
    void add(int t,int v)
    {
    for(int i = n; i >= t; i--)
    dp[i] = max(dp[i],dp[i-t]+v);
    }
    int main()
    {
    while(scanf("%d%d",&n,&m) == 2)
    {
    memset(dp,0,sizeof(dp));
    int a,b;
    for(int i = 0; i < m; i++)
    {
    scanf("%d%d",&a,&b);
    add(a,b);
    }
    printf("%I64d",dp[n]);
    }
    }

  • 0
    @ 2015-08-12 15:28:21

    var f:array[0..1001,0..1001] of integer;
    w,v:array[1..5000] of integer;
    t,m,i,j:integer;
    begin
    readln(t,m);
    for i:=1 to m do read(w[i],v[i]);
    for i:=1 to t do f[0,i]:=0;
    for i:=1 to m do f[i,0]:=0;
    for i:=1 to m do
    for j:=1 to t do begin
    f[i,j]:=f[i-1,j];
    if j>=w[i] then
    if f[i-1,j]<=f[i-1,j-w[i]]+v[i] then f[i,j]:=f[i-1,j-w[i]]+v[i];
    end;
    write(f[m,t]);
    end.

  • 0
    @ 2015-08-11 19:40:41

    program beibao01;
    var
    n,i,mz:longint;
    m,v:array[0..100] of longint;

    function dp(a,b:integer):longint;
    var j:longint;
    begin
    if a=1 then
    begin if b>=m[1] then dp:=v[1]
    else dp:=0;
    end

    else
    begin
    if b>=m[a] then
    begin if dp(a-1,b)>=dp(a-1,b-m[a])+v[a]
    then dp:=dp(a-1,b)

    else dp:=dp(a-1,b-m[a])+v[a] end
    else dp:=dp(a-1,b)
    end;
    end;
    begin
    read(mz,n);
    for i:=1 to n do readln(m[i],v[i]);

    writeln(dp(n,mz));

    end.

  • 0
    @ 2015-08-11 19:38:17

    uses math;
    var
    n,m,v,i,mz,j:longint;

    dp:array[0..100,0..1000] of longint;

    begin
    readln(mz,n);
    for j:=1 to mz do dp[0,j]:=0;
    for i:=1 to n do
    begin
    readln(m,v);
    for j:=m to mz do dp[i,j]:=max(dp[i-1,j],dp[i-1,j-m]+v);
    for j:=1 to m-1 do dp[i,j]:=dp[i-1,j];

    end;
    writeln(dp[n,mz]);
    end.

  • 0
    @ 2015-08-11 19:37:18

    uses math;
    var
    n,m,v,i,mz,j:longint;

    dp:array[0..100,0..1000] of longint;

    begin
    readln(mz,n);
    for j:=1 to mz do dp[0,j]:=0;
    for i:=1 to n do
    begin
    readln(m,v);
    for j:=m to mz do dp[i,j]:=max(dp[i-1,j],dp[i-1,j-m]+v);
    for j:=1 to m-1 do dp[i,j]:=dp[i-1,j];

    end;
    writeln(dp[n,mz]);
    end.

  • 0
    @ 2015-08-11 19:23:40

    program beibao01;
    var
    n,i,mz:longint;
    m,v:array[0..100] of longint;

    function dp(a,b:integer):longint;
    var j:longint;
    begin
    if a=1 then
    begin if b>=m[1] then dp:=v[1]
    else dp:=0;
    end

    else
    begin
    if b>=m[a] then
    begin if dp(a-1,b)>=dp(a-1,b-m[a])+v[a]
    then dp:=dp(a-1,b)

    else dp:=dp(a-1,b-m[a])+v[a] end
    else dp:=dp(a-1,b)
    end;
    end;
    begin
    read(mz,n);
    for i:=1 to n do readln(m[i],v[i]);

    writeln(dp(n,mz));

    end.

  • 0
    @ 2015-08-11 19:23:09

    uses math;
    var
    mz,j:1..1000;
    n,m,v,i:1..100;
    dp:array[0..100,0..1000] of longint;

    begin
    readln(mz,n);
    for j:=1 to mz do dp[0,j]:=0;
    for i:=1 to n do
    begin
    readln(m,v);
    for j:=m to mz do dp[i,j]:=max(dp[i-1,j],dp[i-1,j-m]+v);
    for j:=1 to m-1 do dp[i,j]:=dp[i-1,j];

    end;
    writeln(dp[n,mz]);
    end.

  • 0
    @ 2015-08-11 18:57:49

    program beibao01;
    var
    n,i,mz:longint;
    m,v:array[0..100] of longint;

    function dp(a,b:integer):longint;
    var j:longint;
    begin
    if a=1 then
    begin if b>=m[1] then dp:=v[1]
    else dp:=0;
    end

    else
    begin
    if b>=m[a] then
    begin if dp(a-1,b)>=dp(a-1,b-m[a])+v[a]
    then dp:=dp(a-1,b)

    else dp:=dp(a-1,b-m[a])+v[a] end
    else dp:=dp(a-1,b)
    end;
    end;
    begin
    read(mz,n);
    for i:=1 to n do readln(m[i],v[i]);

    writeln(dp(n,mz));

    end.

  • 0
    @ 2015-08-11 18:40:08

    program beibao01;
    var
    n,i,mz:longint;
    m,v:array[0..100] of longint;

    function dp(a,b:integer):longint;
    var j:longint;
    begin
    if a=1 then
    begin if b>=m[1] then dp(1,b):=v[1]
    else dp(1,b):=0;
    end

    else
    begin
    if b>=m[a] then
    begin if dp(a-1,b)>=dp(a-1,b-m[a])+v[a]
    then dp(a,b):=dp(a-1,b)

    else dp(a,b):=dp(a-1,b-m[a])+v[a] end
    else dp(a,b):=dp(a-1,b)
    end;
    end;
    begin
    read(mz,n);
    for i:=1 to n do readln(m[i],v[i]);

    writeln(dp(n,mz));

    end.

  • 0
    @ 2015-07-15 21:49:14

    #include<iostream>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int MAXN=1000000;

    int t[MAXN],v[MAXN],f[MAXN];

    int main()
    {
    int T,M;
    cin>>T>>M;
    for(int i=1; i<=M; i++)
    cin>>t[i]>>v[i];
    for(int i=1; i<=M; i++)
    for(int j=T; j>=t[i]; j--)
    f[j] = max(f[j],f[j-t[i]]+v[i]);
    cout<<f[T];

    system("pause");
    return 0;
    }

  • 0
    @ 2015-06-06 19:22:26

    var
    a,b,f:array[0..1000]of longint;
    n,i,j,t:longint;
    function max(a,b:longint):longint;
    begin
    if a>b then exit(a) else exit(b);
    end;
    begin
    read(t,n);
    for i:=1 to n do
    read(b[i],a[i]);
    for i:=1 to n do
    for j:=t downto b[i] do
    f[j]:=max(f[j],f[j-b[i]]+a[i]);
    write(f[t]);
    end.

  • 0
    @ 2015-05-28 16:29:01

    第一道自主完成的DP题目...

    #include<stdio.h>
    int main( )
    {
    int t,m,t1[101],p1[101],ans[1001]={0};
    int i,j,k;

    scanf("%d %d",&t,&m);

    for(i=1;i<=m;i++)
    scanf("%d %d",&t1[i],&p1[i]);

    for(i=1;i<=m;i++)
    for(j=t;j>=t1[i];j--)
    if(ans[j-t1[i]]+p1[i]>ans[j]) ans[j]=ans[j-t1[i]]+p1[i];

    printf("%d",ans[t]);

    return 0;

    }

    • @ 2015-05-28 16:32:03

      刚看了下面的题解感觉确实不必用三个数组啊

  • 0
    @ 2015-05-01 20:25:08

    /*************************************************************************
    > File Name: 1104.cpp
    > Author: Netcan
    > Blog: http://www.netcan.xyz
    > Mail: 1469709759@qq.com
    > Created Time: Fri 01 May 2015 20:15:19 CST
    ************************************************************************/

    #include <iostream>
    #include <vector>
    #include <string>
    #include <cstring>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <ctime>
    #include <cstdio>
    #include <sstream>
    #include <deque>
    #include <functional>
    #include <iterator>
    #include <list>
    #include <map>
    #include <memory>
    #include <stack>
    #include <set>
    #include <numeric>
    #include <utility>
    using namespace std;

    int main()
    {
    int T, M;
    int t[105];
    int v[105];
    int dp[105][1005];
    while(scanf("%d%d", &T, &M) == 2) {
    for(int i=1; i<=M; ++i)
    scanf("%d%d", &t[i], &v[i]);
    memset(dp, 0, sizeof(dp));
    for(int i=1;i<=M; ++i)
    for(int j=0; j<=T; ++j) {
    if(j-t[i]>=0) dp[i][j] = max(dp[i-1][j], dp[i-1][j-t[i]]+v[i]);
    else dp[i][j] = dp[i-1][j];
    }
    printf("%d\n", dp[M][T]);
    }

    return 0;
    }

  • 0
    @ 2015-03-18 10:34:42

    #include<stdio.h>
    int dp[101][1001],time[101],p[101];
    int max(int a,int b)
    {
    if(a>b)return a;
    else return b;
    }
    int main()
    {
    int i,j,t,m;
    scanf("%d%d",&t,&m);
    for(i=0;i<m;i++)
    {
    scanf("%d%d",&time[i],&p[i]);
    }
    for(i=m-1;i>=0;i--)
    {
    for(j=0;j<=t;j++)
    {
    dp[i][j]=(i==m-1?0:dp[i+1][j]);
    if(j>=time[i])dp[i][j]=max(dp[i][j],dp[i+1][j-time[i]]+p[i]);
    }
    }
    printf("%d",dp[0][t]);
    return 0;
    }

  • 0
    @ 2015-02-12 16:09:14

    uses math;
    var t,m,i,j:longint;
    a:array[0..100,1..2]of longint;
    f:array[0..10000]of longint;
    procedure init;
    begin
    read(t,m);
    for i:=1 to m do readln(a[i,1],a[i,2]);
    end;
    procedure dp;
    begin
    for i:=1 to m do
    for j:=t downto 0 do
    if (j>=a[i,1]) then f[j]:=max(f[j-a[i,1]]+a[i,2],f[j]);
    write(f[t]);
    end;
    begin
    init;
    dp;
    end.

  • 0
    @ 2015-02-12 16:08:18

    uses math;
    var t,m,i,j:longint;
    a:array[0..100,1..2]of longint;
    f:array[0..10000]of longint;
    procedure init;
    begin
    read(t,m);
    for i:=1 to m do readln(a[i,1],a[i,2]);
    end;
    procedure dp;
    begin
    for i:=1 to m do
    for j:=t downto 0 do
    if (j>=a[i,1]) then f[j]:=max(f[j-a[i,1]]+a[i,2],f[j]);
    write(f[t]);
    end;
    begin
    init;
    dp;
    end.

  • 0
    @ 2015-02-03 15:59:28

    别问我为什么交这么多次,我就是想知道怎么弄才能弄成0ms。who can tell me?

  • 0
    @ 2015-01-17 15:02:31

    #include<iostream>
    using namespace std;
    int w[2000],c[2000];
    int f[2000];
    int main()
    {
    int m,n;
    cin>>m>>n;
    for(int i=1;i<=n;++i)
    cin>>w[i]>>c[i];
    for(int i=1;i<=n;++i)
    for(int v=m;v>=w[i];v--)
    if(f[v-w[i]]+c[i]>f[v])
    f[v]=f[v-w[i]]+c[i];
    cout<<f[m];
    return 0;
    }

信息

ID
1104
难度
4
分类
动态规划 | 背包 点击显示
标签
递交数
16820
已通过
6526
通过率
39%
被复制
38
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