Codeforces 467C George and Job | DP

 

 

 

题目大意：给定一个长度为n的序列，从序列中选出k个不重叠且连续的m个数，要求和最大。
解题思路：dp[i][j]表示到第i个位置选了j个子序列的最大值；可由两个状态转移来：1.保持上一个位置最大值 2.=上一个子序列位置

#include "bits/stdc++.h"
using namespace std;
#define rep(i, s, n) for(int i=s;i<n;i++)
#define LL long long
#define ll __int64
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define E 2.71828
#define MOD 1000000007
#define NN 110

typedef pair<int, int> PII;


const int maxn = 5005;
int N, M, K;
ll arr[maxn], sum[maxn];
ll dp[maxn][maxn];

ll solve () {
    ll ret = 0;
    for (int i = M; i <= N; i++) {
        ll tmp = sum[i] - sum[i-M];
        for (int j = 1; j <= K; j++)
            dp[i][j] = max(dp[i-1][j], dp[i-M][j-1] + tmp);
        ret = max(ret, dp[i][K]);
    }
    /*rep(i,M,N+1){
        rep(j,1,K+1){
            printf("%d ",dp[i][j]);
        }
        printf("\n");
    }*/
    return ret;
}

int main () {
#ifdef ___LOCAL_WONZY___
    freopen("input.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif // ___LOCAL_WONZY___

    while(~scanf("%d%d%d", &N, &M, &K)){
        for (int i = 1; i <= N; i++) {
            scanf("%I64d", &arr[i]);
            sum[i] = sum[i-1] + arr[i];
        }
        printf("%I64d\n", solve());
    }
    return 0;
}

 

参考：http://blog.csdn.net/keshuai19940722/article/details/39585219