HDU1003 最大子段和 线性dp

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): 282195    Accepted Submission(s): 67034

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

 

分析：

题目的数据很水啊

输入6 2 7 -9 5 4 3，答案应该是 12 1 6 的结果12 3 6竟然能过！！！！！！！

ac代码

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
#define max_v 100005
int a[max_v];
int dp[max_v];
int main()
{
    int t;
    cin>>t;
    int c=1;
    while(c<=t)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        //dp[i] 以第i个数结尾的序列的最大字段和
        dp[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            if(dp[i-1]<0)
                dp[i]=a[i];
            else
                dp[i]=dp[i-1]+a[i];
        }
        int index1=1,index2=1;

        //找尾 最大dp[i]对应的i就是尾
        int temp=dp[1];
        for(int i=2;i<=n;i++)
        {
            if(temp<dp[i])
            {
                temp=dp[i];
                index2=i;
            }
        }

        //找头 从尾往前面加，加到和为0就是头
        for(int i=index2,x=0;x!=temp;i--)
        {
            x+=a[i];
            index1=i;
        }
        int sum=0;
        for(int i=index2-1;i>=1;i--)
        {
            sum+=a[i];
            if(sum==0)
                index1=i;
        }

        printf("Case %d:
",c);
        printf("%d %d %d
",temp,index1,index2);
        if(c<t)
            printf("
");
        c++;
    }
    return 0;
}

 另外一种找头的方法：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
#define max_v 100005
int a[max_v];
int dp[max_v];
int main()
{
    int t;
    cin>>t;
    int c=1;
    while(c<=t)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        //dp[i] 以第i个数结尾的序列的最大字段和
        dp[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            if(dp[i-1]<0)
                dp[i]=a[i];
            else
                dp[i]=dp[i-1]+a[i];
        }
        int index1=1,index2=1;

        //找尾 最大dp[i]对应的i就是尾
        int temp=dp[1];
        for(int i=2;i<=n;i++)
        {
            if(temp<dp[i])
            {
                temp=dp[i];
                index2=i;
            }
        }

        index1=index2;
        for(int i=index1;i>=1;i--)
        {
            if(dp[i]>=0)
                index1=i;
            else
                break;
        }

        printf("Case %d:
",c);
        printf("%d %d %d
",temp,index1,index2);
        if(c<t)
            printf("
");
        c++;
    }
    return 0;
}