hdu1003(动态规划，最大子序列和)

Max Sum

Time Limit: 1000 MS Memory Limit: 32768 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

//并查集.cpp
#include
using namespace std;//由于本题从前往后更新，所以不用另开数组
int dp[100005];//dp[i]记录的是以dp[i]为结尾的最大值；
int main()
{
	int n;cin>>n;
	for(int kk=1;kk<=n;kk++){
		int t;cin>>t;
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=t;i++)
		{
			scanf("%d",&dp[i]);
		}
		int l=1,r=1;//记录区间
		int temp=1;//记录起点
		int MAx=dp[1];
		for(int i=2;i<=t;i++)
		{
            if(dp[i-1]>=0)//dp[i-1]大于0，则对dp[i]有贡献
            {
                dp[i]=dp[i-1]+dp[i];
            }else{temp=i;}
            if(dp[i]>MAx)
            {
            	MAx=dp[i];
            	l=temp;
         	    r=i;
            }
		}
		if(kk>=2){printf("\n");}
		printf("Case %d:\n",kk);
		printf("%d %d %d\n",MAx,l,r);
	}
}