LightOJ 1036 - A Refining Company(DP)

来源：互联网发布：竞品分析ppt模板知乎编辑：程序博客网时间：2024/05/21 14:59

题意：镭提炼厂在北边，铀在西边。传送带不可交叉，可从东往西或从南往北传。问最多传送多少矿。

思路：a[][]维护横向即铀矿的前缀和，b[][]维护镭矿前缀和。

dp[i][j] = max(dp[i-1][j] + a[i][j], dp[i][j-1] + b[i][j])。

#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 = 5e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int a[N][N], b[N][N], dp[N][N];int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    sii(n, m);    mem(a, 0), mem(b, 0), mem(dp, 0);    for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) {    si(a[i][j]);    a[i][j] += a[i][j-1];    }    for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) {    si(b[i][j]);    b[i][j] += b[i-1][j];    }    for(int i = 1; i <= n; i ++) {    for(int j = 1; j <= m; j ++) {    dp[i][j] = max(dp[i-1][j] + a[i][j], dp[i][j-1] + b[i][j]);    }    }printf("Case %d: %d\n", ++ cas, dp[n][m]);    }        return 0;}

记忆化搜索：

int dfs(int x, int y) {int& ret = dp[x][y];if(x * y == 0) return 0;if(~ret) return ret;ret = max(ret, dfs(x - 1, y) + a[x][y]);ret = max(ret, dfs(x, y - 1) + b[x][y]);return ret;}

dp初始化为-1。

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