poj 3661 (dp）

The cows are trying to become better athletes, so Bessie is running on a track for exactly N (1 ≤ N ≤ 10,000) minutes. During each minute, she can choose to either run or rest for the whole minute.

The ultimate distance Bessie runs, though, depends on her ‘exhaustion factor’, which starts at 0. When she chooses to run in minute i, she will run exactly a distance of Di (1 ≤ Di ≤ 1,000) and her exhaustion factor will increase by 1 – but must never be allowed to exceed M (1 ≤ M ≤ 500). If she chooses to rest, her exhaustion factor will decrease by 1 for each minute she rests. She cannot commence running again until her exhaustion factor reaches 0. At that point, she can choose to run or rest.

At the end of the N minute workout, Bessie’s exaustion factor must be exactly 0, or she will not have enough energy left for the rest of the day.

Find the maximal distance Bessie can run.

Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 contains the single integer: Di

Output
* Line 1: A single integer representing the largest distance Bessie can run while satisfying the conditions.
　

Sample Input
5 2
5
3
4
2
10
Sample Output
9

题意：N分钟，小刚每分钟可以选择是停下休息还是往前走，如果选择往前走的话可以在第i分钟走Di，当然，小刚的疲劳值也会增加1。如果选择休息则小刚的疲劳值会每分钟减小1，当小刚停下来以后，小刚只能在疲劳值降到0时才能继续再走。已知小刚的疲劳值最多不能超过M，当小刚度过N分钟后要保证他的疲劳值为0，问小刚最多能走多长的距离。

　　分析：先设dp[i][j]表示小明在i分钟，疲劳值为j时所能走的最远距离。

　　a)、先看dp[i][0]的情况，表示第i分钟时，疲劳值为0，考虑这个值由哪些情况得到，1、dp[i][0] = dp[i-1][0]，这个没有任何问题。2、dp[i][0] = dp[i-j][j]。表示i-j分钟时的疲劳值为j，然后一直休息j分钟把疲劳值降成0。

　　b)、现在考虑dp[i][j]的情况，它可以由dp[i-1][j-1] + Di得到，表示第i分钟选择走Di。因为要保证没有后效性，所以只有这一种情况可以转移。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>using namespace std;const int N=10010;const int M=520;int dp[N][M];int d[N];int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {        scanf("%d",&d[i]);    }    dp[1][0]=0;    dp[1][1]=d[1];    for(int i=2;i<=n;i++)    {        dp[i][0]=dp[i-1][0];        for(int j=1;j<=m;j++)        {            if(i-j>=0)            dp[i][0]=max(dp[i][0],dp[i-j][j]);//一开始的想法是 dp[i][j]=max(dp[i-1][j-1],dp[i-1][j+1]),        //但是没考虑周到的是 一旦他选择休息，必定休息到为0，所以不是        //任何疲劳值都能休息，必然是(i-j>=0) 转移，dp[i][0]=max(dp[i-1][0],dp[i-j][j]);        ////就是在i-j阶段的时候选择休息，而且疲劳值为j，休息了j分钟，疲劳为0.        dp[i][j]=dp[i-1][j-1]+d[i];        }    }    printf("%d\n",dp[n][0] );}