Maximum sum-动态规划

A - Maximum sum

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2479

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

Your task is to calculate d(A).

Input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input. 
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer. 

Huge input,scanf is recommended.

/*
Author: 2486
Memory: 544 KB		Time: 438 MS
Language: C++		Result: Accepted
*/
//题目思路是从左到右分别求出它们所在位置的最大连续和，然后从右到左求出它们所在的最大连续和
//接着就是a[i]+b[i+1],a数组代表着从左到右,b代表着从右到左所以不断的比較a[0]+b[1],a[1]+b[2]求最大值就可以
//如何求解最大值（代码最后段有具体解释）
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=50000+5;
int n,m;
int a[maxn];
int dp[maxn];
int main() {
    scanf("%d",&n);
    while(n--) {
        scanf("%d",&m);
        for(int i=0; i<m; i++) {
            scanf("%d",&a[i]);
        }
        dp[0]=a[0];
        int sum=a[0];
        int ans=a[0];
        for(int i=1; i<m; i++) {
            if(sum<0) {
                sum=0;
            }
            sum+=a[i];
            if(sum>ans) {
                ans=sum;
            }
            dp[i]=ans;
        }
        sum=a[m-1];
        int Max=dp[m-2]+sum;
        ans=a[m-1];
        for(int j=m-2; j>=1; j--) {
            if(sum<0) {
                sum=0;
            }
            sum+=a[j];
            if(sum>ans) {
                ans=sum;
            }
            Max=max(Max,dp[j-1]+ans);
        }
        printf("%d
",Max);
    }


    return 0;

}

假设当前的数据和小于零。非常明显，将它增加到后面的计算中，肯定会降低最大值
非常easy的道理。-1+4<0+4，假设之前的取值小于零，抛弃它，又一次赋值为零
然后通过maxs不断更新当前的最大值
while(true){
            scanf("%d",&a);
            if(sum<0) {
                sum=0;
            }
            sum+=a;
            if(sum>maxs) {
                maxs=sum;
            }
}