HDU-1003 Max Sum （DP）

Max Sum

http://acm.hdu.edu.cn/showproblem.php?pid=1003

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

   
   
   
   

    
    
    
    2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
14 1 4

Case 2:
7 1 6
   
   
   
   

题目大意：给定一串数字，求满足子串和最大的子串及其起始和结束下标？

大致思路：定义dp[i]表示以i为结尾的子串的最大和，再用sta数组保存其起始下标即可，进行状态转移

刚开始对sta数组也开了一个数组存储下标，但是发现cur只用一次，如果及时更新答案就不用数组保存了，然后就换成curSta表示当前子串的起始下标（其实dp数组也可以不需要，但是为了符合习惯，就不省略了）

过了这么久再写一次，发现写的方法还是一样（总想着减少时间和空间）

#include <cstdio>

using namespace std;

int dp[100005],curSta,ansSum,ansSta,ansDes;//dp[i]表示以i为结尾的子串的最大和，curSta表示其起始下标
int n;

int main() {
    int T,kase=0;
    scanf("%d",&T);
    while(kase<T) {
        if(kase!=0)
            printf("\n");
        printf("Case %d:\n",++kase);
        scanf("%d%d",&n,dp+1);
        curSta=ansSta=ansDes=1;
        ansSum=dp[1];
        for(int i=2;i<=n;++i) {
            scanf("%d",dp+i);
            if(dp[i-1]<0) {
                curSta=i;
            }
            else {
                dp[i]+=dp[i-1];
            }
            if(ansSum<dp[i]) {
                ansSum=dp[i];
                ansSta=curSta;
                ansDes=i;
            }
        }
        printf("%d %d %d\n",ansSum,ansSta,ansDes);
    }
    return 0;
}