hdu 1003 Max Sum（基础dp）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72615    Accepted Submission(s): 16626

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

 

算法分析：求最大字段和，d[i]表示已 i 结尾（字段和中包含 i ）在 a[1..i] 上的最大和，d[i]=(d[i-1]+a[i]>a[i])?d[i-1]+a[i]:a[i];

max = {d[i],1<=i<=n} ;

#include<iostream>
#define N 100010
using namespace std;
int a[N],d[N];
int main()
{
    int test,n,i,max,k,f,e;
    cin>>test;
    k=1;
    while(test--)
    {
        cin>>n;
        for(i=1;i<=n;i++)
            cin>>a[i];
        d[1]=a[1];
        for(i=2;i<=n;i++)
        {
            if(d[i-1]<0) d[i]=a[i];
            else d[i]=d[i-1]+a[i];
        }
        max=d[1];e=1;
        for(i=2;i<=n;i++)
        {
            if(max<d[i])
            {
                max=d[i];e=i;
            }
        }
        int t=0;
        f=e;
        for(i=e;i>0;i--)
        {
            t=t+a[i];
            if(t==max)    f=i;
        }
        cout<<"Case "<<k++<<":"<<endl<<max<<" "<<f<<" "<<e<<endl;
        if(test) cout<<endl;
    }
    return 0;
}

改进后的只处理最大和不处理位置

#include<cstdio>
int main()
{
    int n,test,ans,t,a,i;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d",&n);
        scanf("%d",&a);
        ans=t=a;
        for(i=1;i<n;i++) 
        {
            scanf("%d",&a);
            if(t<0) t=a;
            else t=t+a;
            if(ans<t) ans=t;
        }
        printf("%d
",ans);
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int T;
    scanf("%d", &T);

    int N;
    int a;
    int ans;
    int sum;
    int i;
    int bg, ed;//起始，结束
    int bg2;
    int cas = 0;

    while (T--) {
        scanf("%d", &N);

        ans = -1010;//
        sum = 0;//
        bg2 = 0;//默认起始位置
        for (i = 0; i < N; ++i) {
            scanf("%d", &a);

            sum = sum + a;
            if (sum > ans) {
                ans = sum;
                bg = bg2;
                ed = i;
            }
            if (sum < 0) {//< 0 那么这段到此为止吧
                sum = 0;//
                bg2 = i + 1;//更新起始位置
            }
        }

        printf("Case %d:
", ++cas);
        printf("%d %d %d
", ans, bg + 1, ed + 1);
        if (T > 0) {
            printf("
");
        }
    }
    return 0;
}

这个和上面的相比写法容易理解些

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int T;
    scanf("%d", &T);

    int N;
    int a;
    int ans;
    int sum;
    int i;
    int bg, ed;//起始，结束
    int bg2;
    int cas = 0;

    while (T--) {
        scanf("%d", &N);

        scanf("%d", &a);
        ans = a;
        sum = a;
        bg = 0, ed = 0;
        bg2 = 0;

        for (i = 1; i < N; ++i) {
            scanf("%d", &a);

            if (sum <= 0) {//从样例2 看，这里要<，但是<= 也可以，只要找到一个最大的子串就可以
                sum = a;
                bg2 = i;
            } else {
                sum = sum + a;
            }

            if (sum >= ans) {//这里> 和>= 都可以
                ans = sum;
                bg = bg2, ed = i;
            }
        }

        printf("Case %d:
", ++cas);
        printf("%d %d %d
", ans, bg + 1, ed + 1);
        if (T > 0) {
            printf("
");
        }
    }
    return 0;
}