hdu 1559【最大子矩陣和DP】

這題比poj那道多了限制，就是規定了子矩陣大小，開始看到時間10s，果斷暴力，接著又果斷TLE，加一點點最佳化還是TLE，坑爹！

接著就用DP，怎麼DP？很簡單，令a[i][j]為1<=s<=i，1<=t<=j所有元素的和，可以線上處理，那麼就可以通過加減來求出當前子矩陣和，就好像計算幾個矩陣面積一樣（某部分覆蓋在一起，所以相加後減去重複的面積）。

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a>b?b:a;}#define LL long long#define PI acos(-1.0)#define N  1010#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)int n,m,x,y;int a[N][N];int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d%d",&n,&m,&x,&y);        int i,j;        int ans=-1;        memset(a,0,sizeof(a));        for(i=1;i<=n;i++)        {            for(j=1;j<=m;j++)            {                scanf("%d",&a[i][j]);                a[i][j]+=a[i-1][j]+a[i][j-1]-a[i-1][j-1];                if(i>=x && j>=y)                {                    int tmp=a[i][j]-a[i-x][j]-a[i][j-y]+a[i-x][j-y];                    if(tmp>ans)ans=tmp;                }            }        }        printf("%d\n",ans);    }    return 0;}