51nod 1052最大M子段和 & poj 2479最大两子段和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <limits.h>
using namespace std;
typedef long long LL;
#define MAXN 5010
const LL MIN_INF = -(1 << 30);
LL dp[MAXN], input[MAXN], pre[MAXN];
int n, m;
int main() {
    scanf("%d%d", &n, &m);
    dp[0] = MIN_INF;
    pre[0] = 0;
    for(int i = 1; i <= n; i++) {
        scanf("%I64d", &input[i]);
        pre[i] = 0;
        dp[i] = MIN_INF;
    }
    LL ans = MIN_INF;
    while(m--) {
        ans = MIN_INF;
        for(int i = 1; i <= n; i++) {
            if(i == 1) dp[i] = pre[i - 1] + input[i];
            else {
                if(dp[i - 1] > pre[i - 1]) {
                    dp[i] = dp[i - 1] + input[i];//第i个数与它的上一个数在同一子段内
                } else {
                    dp[i] = pre[i - 1] + input[i];//从第i个分出一个新的子段
                }
            }
            //dp[i] = max(dp[i - 1], pre[i - 1]) + input[i]; 或者直接这么取值也可
            pre[i - 1] = ans;
            ans = max(ans, dp[i]);
        }
        pre[n] = ans;
    }
    printf("%I64d\n", pre[n]);
    return 0;
}