标签：dp mss
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158875 Accepted Submission(s): 37166
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
Author
Ignatius.L
题意: 求最大字段和，并给出字段的起始点和终点。
题解:DP之，模板题——O(n)的算法。
AC代码:
#include<iostream>
#define N 100000+5
using namespace std;
int dp[N],t,n;
int main()
{
cin.sync_with_stdio(false);
cin>>t;
for(int k=1;k<=t;k++){
cin>>n;
for(int i=0;i<n;i++)cin>>dp[i];
int sum=dp[0],tmp=dp[0],s=1,b=1,d=1;
for(int i=1;i<n;i++){
if(tmp>=0)
tmp+=dp[i];
else
tmp=dp[i],s=i+1;
if(tmp>sum) sum=tmp,d=i+1,b=s;
}
cout<<"Case "<<k<<":"<<endl;
cout<<sum<<" "<<b<<" "<<d<<endl;
if(k!=t)
cout<<endl;
}
return 0;
}
HDOJ 1003 Max Sum【MSS】
标签：dp mss
原文地址：http://blog.csdn.net/mummyding/article/details/43604337
