HDU 1003 Max Sum (动态规划最大区间和）

逃离我推掉我的手 2021-10-30 04:46 279阅读 0赞

# Max Sum #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 330535 Accepted Submission(s): 78678**

### 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

### 题目大意与分析 ###

就是给你一个序列，求最大区间和以及该最大区间的下标

动态规划的思想：

用dp\[i\]来代表以a\[i\]为结尾的最大区间和，这样就有两种情况：

要么dp\[i\]是正的 那么以a\[i\]结尾的最大区间和就是dp\[i-1\]+a\[i\]（因为必须是a\[i\]结尾，a\[i\]是正是负都无法取舍的）

要么dp\[i\]是负的 那么以a\[i\]结尾的最大区间和就是a\[i\]

状态转移方程：dp\[i\] = max( dp\[i-1\]+a\[i\], a\[i\] );

初始化：因为我们从前往后推，可以将dp数组和a数组合二为一，节约空间

另外需要注意的是 需要三个游标：s、l、r 分别代表当前区间的左端点、最大区间的左端点、最大区间的右断点，至于当前区间的右端点，可以直接用i表示。

### 代码 ###

#include<bits/stdc++.h>
    
    using namespace std;
    
    int T,n,i,dp[100005],l,r,s,maxs,t=0;
    
    int main()
    {
        cin>>T;
        while(T--)
        {
            t++;
            memset(dp,0,sizeof(dp));
            cin>>n;
            for(i=1;i<=n;i++)
            {
                scanf("%d",&dp[i]);
            }
            maxs=dp[1];
            s=1;
            l=1;
            r=1;
            for(i=2;i<=n;i++)
            {
                dp[i]=max(dp[i-1]+dp[i],dp[i]);
                if(dp[i-1]<0)
                {
                    s=i;
                }
                if(dp[i]>maxs)
                {
                    l=s;
                    r=i;
                    maxs=dp[i];
                }
            }
            printf("Case %d:\n",t);
            printf("%d %d %d\n",maxs,l,r);    
            if(T)
            printf("\n");
        }
    }

转载于:https://www.cnblogs.com/dyhaohaoxuexi/p/11336609.html