HDU 5569 matrix(DP)

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题意：nxm矩阵，从(1, 1)走到(n, m)，只能往下或往右，求最小的 a1 * a2 + a3 * a4 + ... + a(n+m-2) * a(n+m-1)。a的下标k为第k步走的格子的值。

思路：偶数步加积，奇数步的话就按上一步的值算。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e3 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int dp[N][N], a[N][N];int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//  freopen("/Users/apple/out.txt", "w", stdout);#endifmem(dp, 0x3f);    while(sii(n, m) != EOF) {    for(int i = 1; i <= n; i ++)     for(int j = 1; j <= m; j ++)    si(a[i][j]);    dp[1][0] = dp[0][1] = 0;    for(int i = 1; i <= n; i ++) {    for(int j = 1; j <= m; j ++) {    if((i + j) & 1) dp[i][j] = min(dp[i-1][j] + a[i-1][j] * a[i][j],     dp[i][j-1] + a[i][j-1] * a[i][j]);    else dp[i][j] = min(dp[i-1][j], dp[i][j-1]);    }    }    printf("%d\n", dp[n][m]);    }    return 0;}

记忆化搜索：

int dfs(int x, int y) {int& ret = dp[x][y];if(ret < inf) return ret;if(x == 1 && y == 1) return ret = 0;if(x > 1) {if((x + y) & 1) ret = min(ret, dfs(x - 1, y) + a[x][y] * a[x-1][y]);else ret = min(ret, dfs(x - 1, y));}if(y > 1) {if((x + y) & 1) ret = min(ret, dfs(x, y - 1) + a[x][y] * a[x][y-1]);else ret = min(ret, dfs(x, y - 1));}return ret;}

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