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
, 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

 

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;
}