bzoj 3111: [Zjoi2013]蚂蚁寻路 动态规划

       首先枚举下边界，然后令f[k][x][y]表示已经左转k次，当前在x,y的最优值，转移可以优化到O(1)。因此为O(N^3K)。

AC代码如下：

#include<iostream>#include<cstdio>#include<cstring>#define inf 1000000000#define N 105using namespace std;int m,n,cnt,a[N][N],b[N][N],f[N][N],g[N][N];int main(){scanf("%d%d%d",&m,&n,&cnt); int i,j,k,x,y,ans=-inf;for (i=1; i<=m; i++)for (j=1; j<=n; j++) scanf("%d",&a[i][j]);for (i=1; i<=m; i++){for (x=i; x; x--)for (y=1; y<=n; y++) b[x][y]=b[x+1][y]+a[x][y];memset(f,200,sizeof(f)); memset(g,200,sizeof(g));for (x=1; x<=i; x++)for (y=1; y<=n; y++) g[x][y]=max(g[x][y-1],0)+b[x][y];for (j=1; j<=cnt; j++){for (y=1; y<=n; y++)for (k=-inf,x=1; x<=i; x++){k=max(k,g[x-1][y-1]);f[x][y]=max(k,f[x][y-1])+b[x][y];}for (y=1; y<=n; y++)for (k=-inf,x=i; x; x--){k=max(k,f[x+1][y-1]);g[x][y]=max(k,g[x][y-1])+b[x][y];}}for (x=1; x<=i; x++)for (y=1; y<=n; y++) ans=max(ans,g[x][y]);}printf("%d\n",ans);return 0;}

by lych

2016.5.16