poj 3920(简单dp)

来源：互联网发布：英语短语软件知乎编辑：程序博客网时间：2024/05/29 18:18

题意：从起点每次最多跳s步，最多跳t次，从起点到终点的最大得分。

解题思路：dp[i][j]表示第i次跳到位置j的最大得分。简单的动态规划。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 205;const int inf = 0x7ffffff;int n,s,t;int grid[maxn],dp[maxn][maxn];int main(){while(scanf("%d",&n)!=EOF){if(n == 0) break;scanf("%d%d",&s,&t);for(int i = 0; i <= t; i++)for(int j = 0; j <= n+1; j++)dp[i][j] = -inf;memset(grid,0,sizeof(grid));for(int i = 1; i <= n; i++)scanf("%d",&grid[i]);for(int i = 1; i <= s; i++)dp[1][i] = grid[i];for(int i = 1; i < t; i++)for(int j = 1; j <= n; j++){if(dp[i][j] == -1) continue;for(int k = 1; k <= s; k++){int len = j + k; if(len > n) len = n + 1;dp[i+1][len] = max(dp[i+1][len],dp[i][j] + grid[len]);}}int ans = -inf;for(int i = 1; i <= t; i++)ans = max(ans,dp[i][n+1]);printf("%d\n",ans);}return 0;}

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