HDU1003 最大子段和 线性dp

时间：2022-03-13 16:03:45

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[]= {,,,,,,,,,,,,};

int mon2[]= {,,,,,,,,,,,,};

int dir[][]={{,},{,-},{,},{-,}};

#define max_v 100005

int a[max_v];

int dp[max_v];

int main()

{

    int t;

    cin>>t;

    int c=;

    while(c<=t)

    {

        int n;

        scanf("%d",&n);

        for(int i=;i<=n;i++)

            scanf("%d",&a[i]);

        memset(dp,,sizeof(dp));

        //dp[i] 以第i个数结尾的序列的最大字段和

        dp[]=a[];

        for(int i=;i<=n;i++)

        {

            if(dp[i-]<)

                dp[i]=a[i];

            else

                dp[i]=dp[i-]+a[i];

        }

        int index1=,index2=;

        //找尾 最大dp[i]对应的i就是尾

        int temp=dp[];

        for(int i=;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=;x!=temp;i--)

        {

            x+=a[i];

            index1=i;

        }

        int sum=;

        for(int i=index2-;i>=;i--)

        {

            sum+=a[i];

            if(sum==)

                index1=i;

        }

        printf("Case %d:\n",c);

        printf("%d %d %d\n",temp,index1,index2);

        if(c<t)

            printf("\n");

        c++;

    }

    return ;

}

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[]= {,,,,,,,,,,,,};

int mon2[]= {,,,,,,,,,,,,};

int dir[][]={{,},{,-},{,},{-,}};

#define max_v 100005

int a[max_v];

int dp[max_v];

int main()

{

    int t;

    cin>>t;

    int c=;

    while(c<=t)

    {

        int n;

        scanf("%d",&n);

        for(int i=;i<=n;i++)

            scanf("%d",&a[i]);

        memset(dp,,sizeof(dp));

        //dp[i] 以第i个数结尾的序列的最大字段和

        dp[]=a[];

        for(int i=;i<=n;i++)

        {

            if(dp[i-]<)

                dp[i]=a[i];

            else

                dp[i]=dp[i-]+a[i];

        }

        int index1=,index2=;

        //找尾 最大dp[i]对应的i就是尾

        int temp=dp[];

        for(int i=;i<=n;i++)

        {

            if(temp<dp[i])

            {

                temp=dp[i];

                index2=i;

            }

        }

        index1=index2;

        for(int i=index1;i>=;i--)

        {

            if(dp[i]>=)

                index1=i;

            else

                break;

        }

        printf("Case %d:\n",c);

        printf("%d %d %d\n",temp,index1,index2);

        if(c<t)

            printf("\n");

        c++;

    }

    return ;

}

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