hdu5569 matrix

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思路：dp[i][j]表示到这里的最小和。然后就是分奇偶讨论了。

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>// #define DEBUG#ifdef DEBUG#define debug(...) printf( __VA_ARGS__ )#else#define debug(...)#endif#define MEM(x,y) memset(x, y,sizeof x)using namespace std;typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const LL inf = 1LL << 60;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int n, m;int A[1010][1010];int step[110][110];LL dp[1010][1010];int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(scanf("%d%d",&n,&m) != EOF){for (int i = 1;i <= n;++i)for (int j = 1;j <= m;++j)scanf("%d",&A[i][j]);for (int i = 0;i <= n + 1;++i)for (int j = 0;j <= m + 1;++j)dp[i][j] = inf;dp[0][1] = dp[1][0] = 0;dp[1][1] = 0;for (int i = 1;i <= n;++i){for (int j = 1;j <= m;++j){if ((i+j)%2){dp[i][j] = min(dp[i-1][j] + 1LL*A[i-1][j]*A[i][j], dp[i][j]);dp[i][j] = min(dp[i][j-1] + 1LL*A[i][j-1]*A[i][j], dp[i][j]);}else dp[i][j] = min(dp[i][j-1], dp[i-1][j]);}}printf("%lld\n", dp[n][m]);}return 0;}

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