[POJ 2444] Partition a Matrix 暴力

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http://poj.org/problem?id=2444

题意：用两条线将矩阵切成三部分，两天直线可以平行也可以垂直，如果垂直的话有一条就是射线（要不然变成四部分了）

思路：情况不多暴力枚举。。。。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;typedef __int64 ll;const ll inf = 0x7fffffffffffffff;ll minn;int n, m;ll dp[505][505];int mapn[505][505];inline ll abs(ll a){    return a >= 0 ? a : -a;}void Init()  //初始化，求(1, 1)到（i，k）矩阵的和{    for(int i = 0; i <= n || i <= m; i++){        dp[i][0] = dp[0][i] = 0;    }    for(int i = 1; i <= n; i++){        for(int k = 1; k <= m; k++){            dp[i][k] =dp[i-1][k] + dp[i][k-1] - dp[i-1][k-1] + mapn[i][k];        }    }}inline ll fun(int up, int dw, int lf, int rg) // 求和{    ll sum = 0;    sum += dp[dw][rg] + dp[up-1][lf-1] - dp[up-1][rg] - dp[dw][lf-1];    return sum;}inline void fuck(ll sum1, ll sum2, ll sum3){    ll sum = abs(sum1 - sum2) + abs(sum1 - sum3) + abs(sum2 - sum3);    if(sum < minn)        minn = sum;}int main(){    while(scanf("%d%d", &n, &m) && (m || n)){        for(int i = 1; i <= n; i++){            for(int k = 1; k <= m; k++){                scanf("%d", &mapn[i][k]);            }        }        Init();        minn = inf;        for(int i = 1; i < n; i++){ //垂直情况            for(int k = 1; k < m; k++){                ll sum1 = fun(1, i, 1, k);                ll sum2 = fun(1, i, k+1, m);                ll sum3 = fun(i+1, n, 1, k);                ll sum4 = fun(i+1, n, k+1, m);                fuck(sum1+sum2, sum3, sum4);                fuck(sum1+sum3, sum2, sum4);                fuck(sum2+sum4, sum1, sum3);                fuck(sum3+sum4, sum1, sum2);            }        }        for(int i = 1; i < n-1; i++){ //平行                ll sum1 = fun(1, i, 1, m);            for(int k = i+1; k < n; k++){                ll sum2 = fun(i+1, k, 1, m);                ll sum3 = fun(k+1, n, 1, m);                fuck(sum1, sum2, sum3);            }        }        for(int i = 1; i < m-1; i++){//平行                ll sum1 = fun(1, n, 1, i);            for(int k = i+1; k < m; k++){                ll sum2 = fun(1, n, i+1, k);                ll sum3 = fun(1, n, k+1, m);                fuck(sum1, sum2, sum3);            }        }        printf("%I64d\n", minn);    }    return 0;}

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