poj 3920(简单dp)

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;
}