题解

3 条题解

  • 0
    @ 2023-07-15 12:47:14
    #include <queue>
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cmath>
    
    using namespace std;
    #define Rep(i,_z,_zz) for(int i = (_z); i <= (_zz); i++)
    #define Red(i,_z,_zz) for(int i = (_z); i >= (_zz); i--)
    
    inline void read(int &x) {
        register char c;
        while((c=getchar()) < '0' || c  > '9');
        x = c-'0';
        while((c=getchar()) >= '0' && c <= '9')
          x = x*10+c-'0';
    }
    
    #define N 1510
    int n,m,k,a[N][N],s[N][N],m1[N][N],m2[N][N],m3[N][N],m4[N][N],ans = 0;
    
    inline int Ma(int a,int b){return a>b?a:b;}
    
    void out(int k[N][N]) {
        for (int i = 1; i <= n; i++) {
          for (int j = 1; j <= m; j++)
            printf("%3d",k[i][j]<0?-1:k[i][j]);
          cout << "\n";
        }
    }
    
    int main() {
        //freopen("large-4.in","r",stdin);
        //freopen("oil.out","w",stdout);
        read(n); read(m); read(k);
        memset(m1,-127/3,sizeof m1);
        memset(m2,-127/3,sizeof m2);
        memset(m3,-127/3,sizeof m3);
        memset(m4,-127/3,sizeof m4);
        memset(s,-127/3,sizeof s);
        Rep(i,1,n) Rep(j,1,m) read(a[i][j]),a[i][j] += (a[i-1][j]+a[i][j-1]-a[i-1][j-1]);
        Rep(i,k,n) Rep(j,k,m) s[i][j] = a[i][j]-a[i-k][j]-a[i][j-k]+a[i-k][j-k];
        Rep(i,k,n) Rep(j,k,m) m1[i][j] = Ma(s[i][j],Ma(m1[i-1][j],m1[i][j-1]));
        Rep(i,k,n) Red(j,m-k+1,1) m2[i][j] = Ma(s[i][j+k-1],Ma(m2[i-1][j],m2[i][j+1]));
        Red(i,n-k+1,1) Rep(j,k,m) m3[i][j] = Ma(s[i+k-1][j],Ma(m3[i+1][j],m3[i][j-1]));
        Red(i,n-k+1,1) Red(j,m-k+1,1) m4[i][j] = Ma(s[i+k-1][j+k-1],Ma(m4[i+1][j],m4[i][j+1]));
        //out(m1);
        Rep(i,k,n-k+1) Rep(j,k,m-k+1) {
          ans = Ma(ans,m1[i][j]+m2[i][j+1]+m3[i+1][m]);
          ans = Ma(ans,m3[i][j]+m4[i][j+1]+m1[i-1][m]);
          ans = Ma(ans,m1[i][j]+m3[i+1][j]+m2[n][j+1]);
          ans = Ma(ans,m2[i][j]+m4[i+1][j]+m1[n][j-1]);
          ans = Ma(ans,s[i][j]+m1[n][j-k]+m2[n][j+1]);
          ans = Ma(ans,s[i][j]+m1[i-k][m]+m3[i+1][m]);
        }
        printf("%d",ans);
       // while(1);
        return 0;
    }
    
    
  • -1
    @ 2017-10-09 19:00:54
    #include <queue>
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cmath>
    
    using namespace std;
    #define Rep(i,_z,_zz) for(int i = (_z); i <= (_zz); i++)
    #define Red(i,_z,_zz) for(int i = (_z); i >= (_zz); i--)
    
    inline void read(int &x) {
        register char c;
        while((c=getchar()) < '0' || c  > '9');
        x = c-'0';
        while((c=getchar()) >= '0' && c <= '9')
          x = x*10+c-'0';
    }
    
    #define N 1510
    int n,m,k,a[N][N],s[N][N],m1[N][N],m2[N][N],m3[N][N],m4[N][N],ans = 0;
    
    inline int Ma(int a,int b){return a>b?a:b;}
    
    void out(int k[N][N]) {
        for (int i = 1; i <= n; i++) {
          for (int j = 1; j <= m; j++)
            printf("%3d",k[i][j]<0?-1:k[i][j]);
          cout << "\n";
        }
    }
    
    int main() {
        //freopen("large-4.in","r",stdin);
        //freopen("oil.out","w",stdout);
        read(n); read(m); read(k);
        memset(m1,-127/3,sizeof m1);
        memset(m2,-127/3,sizeof m2);
        memset(m3,-127/3,sizeof m3);
        memset(m4,-127/3,sizeof m4);
        memset(s,-127/3,sizeof s);
        Rep(i,1,n) Rep(j,1,m) read(a[i][j]),a[i][j] += (a[i-1][j]+a[i][j-1]-a[i-1][j-1]);
        Rep(i,k,n) Rep(j,k,m) s[i][j] = a[i][j]-a[i-k][j]-a[i][j-k]+a[i-k][j-k];
        Rep(i,k,n) Rep(j,k,m) m1[i][j] = Ma(s[i][j],Ma(m1[i-1][j],m1[i][j-1]));
        Rep(i,k,n) Red(j,m-k+1,1) m2[i][j] = Ma(s[i][j+k-1],Ma(m2[i-1][j],m2[i][j+1]));
        Red(i,n-k+1,1) Rep(j,k,m) m3[i][j] = Ma(s[i+k-1][j],Ma(m3[i+1][j],m3[i][j-1]));
        Red(i,n-k+1,1) Red(j,m-k+1,1) m4[i][j] = Ma(s[i+k-1][j+k-1],Ma(m4[i+1][j],m4[i][j+1]));
        //out(m1);
        Rep(i,k,n-k+1) Rep(j,k,m-k+1) {
          ans = Ma(ans,m1[i][j]+m2[i][j+1]+m3[i+1][m]);
          ans = Ma(ans,m3[i][j]+m4[i][j+1]+m1[i-1][m]);
          ans = Ma(ans,m1[i][j]+m3[i+1][j]+m2[n][j+1]);
          ans = Ma(ans,m2[i][j]+m4[i+1][j]+m1[n][j-1]);
          ans = Ma(ans,s[i][j]+m1[n][j-k]+m2[n][j+1]);
          ans = Ma(ans,s[i][j]+m1[i-k][m]+m3[i+1][m]);
        }
        printf("%d",ans);
       // while(1);
        return 0;
    }
    
  • -1
    @ 2017-01-13 08:20:21

    uses math;
    var
    a,b,c,d,f:array [0..2505,0..2505] of longint;
    n,m,k,i,j,ans:longint;
    begin
    read(n,m,k);
    for i:=1 to n do
    for j:=1 to m do read(a[i,j]);
    for i:=1 to n do
    for j:=1 to m do a[i,j]:=a[i,j]+a[i,j-1];
    for i:=1 to n do
    for j:=1 to m do a[i,j]:=a[i,j]+a[i-1,j];
    n:=n-k+1; m:=m-k+1;
    for i:=1 to n do
    for j:=1 to m do f[i,j]:=a[i+k-1,j+k-1]-a[i-1,j+k-1]-a[i+k-1,j-1]+a[i-1,j-1];
    fillchar(a,sizeof(a),0);
    for i:=1 to n do
    for j:=1 to m do a[i,j]:=max(f[i,j],max(a[i,j-1],a[i-1,j]));
    for i:=1 to n do
    for j:=m downto 1 do b[i,j]:=max(f[i,j],max(b[i,j+1],b[i-1,j]));
    for i:=n downto 1 do
    for j:=1 to m do c[i,j]:=max(f[i,j],max(c[i,j-1],c[i+1,j]));
    for i:=n downto 1 do
    for j:=m downto 1 do d[i,j]:=max(f[i,j],max(d[i,j+1],d[i+1,j]));
    ans:=0;
    for i:=k+1 to n do
    for j:=k+1 to m do ans:=max(ans,max(a[i-k,j-k]+b[i-k,j]+d[i,1],b[i-k,1]+c[i,j-k]+d[i,j]));
    for i:=k+1 to n do
    for j:=k+1 to m do ans:=max(ans,max(c[1,j-k]+b[i-k,j]+d[i,j],d[1,j]+a[i-k,j-k]+c[i,j-k]));
    for i:=1 to n do
    for j:=k+1 to m-k do ans:=max(ans,a[n,j-k]+f[i,j]+d[1,j+k]);
    for i:=k+1 to n-k do
    for j:=1 to m do ans:=max(ans,a[i-k,m]+f[i,j]+d[i+k,1]);
    write(ans);
    end.

  • 1

信息

ID
2002
难度
5
分类
动态规划 点击显示
标签
递交数
109
已通过
43
通过率
39%
被复制
3
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