HDU：1003 Max Sum（动态规划DP）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 217253    Accepted Submission(s): 51248

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

Author

Ignatius.L

 

Recommend

题目大意：给你一个序列，让你输出这个序列最大子序列之和，再输出这个最大子序列的首尾元素的标号。

解题思路：与上一篇题目差不多一样，只不过细节不大一样，在代码里已经标注出来了，状态转移方程依然为dp[i]=max(a[i],dp[i-1]+a[i])。

代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <queue>#include <stack>#include <algorithm>using namespace std;struct node{int p;int i,j;}dp[100010];int a[100010];bool cmp(struct node a,struct node b){if(a.p==b.p){if(a.i==b.i){return a.j<b.j;}else{return a.i<b.i;}}else{return a.p>b.p;}}int main(){int t;scanf("%d",&t);int hao=1;while(t--){memset(dp,0,sizeof(dp));int n;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}dp[1].p=a[1];dp[1].i=1;dp[1].j=1;for(int i=2;i<=n;i++){if(a[i]>dp[i-1].p+a[i])//7 0 6 -1 1 -6 7 -5，通过这个例子输出7 1 6可以看出 ，这里不加=号 {dp[i].p=a[i];dp[i].i=i;dp[i].j=i;}else{dp[i].p=dp[i-1].p+a[i];dp[i].i=dp[i-1].i;dp[i].j=i;}}sort(dp+1,dp+n+1,cmp);if(hao!=1)//题目中要求的格式，得有一个空行 {printf("\n");}printf("Case %d:\n",hao++);printf("%d %d %d\n",dp[1].p,dp[1].i,dp[1].j);}return 0;}