Max Sum （hdu 1003 简单DP水过）

时间：2023-12-11 08:41:38

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 161294    Accepted Submission(s): 37775

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
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题意：给出n个数的序列，求出最大的子串和。并输出起点和终点。

思路：dp[i]表示以i为结尾的最大子串和。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <vector>

#include <set>

#include <queue>

#pragma comment (linker,"/STACK:102400000,102400000")

#define maxn 100005

#define MAXN 2005

#define mod 1000000009

#define INF 0x3f3f3f3f

#define pi acos(-1.0)

#define eps 1e-6

#define lson rt<<1,l,mid

#define rson rt<<1|1,mid+1,r

#define FRE(i,a,b)  for(i = a; i <= b; i++)

#define FRL(i,a,b)  for(i = a; i < b; i++)

#define mem(t, v)   memset ((t) , v, sizeof(t))

#define sf(n)       scanf("%d", &n)

#define sff(a,b)    scanf("%d %d", &a, &b)

#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)

#define pf          printf

#define DBG         pf("Hi\n")

typedef long long ll;

using namespace std;

int dp[maxn];       //dp[i]表示以i为结尾的最大子串和

int a[maxn];

int n;

int main()

{

    int i,j,t,cas=1;;

    sf(t);

    bool flag=false;

    while (t--)

    {

        if (flag) pf("\n");

        flag=true;

        sf(n);

        FRL(i,1,n+1)

            sf(a[i]);

        mem(dp,0);

        dp[1]=a[1];

        int S=1,T=1,s=1,t=1,maxx=a[1];//s,t记录当前首尾指针。S。T记录当前最大值的首尾指针

        FRL(i,2,n+1)

        {

            if (a[i]>dp[i-1]+a[i])  //假设a[i]对dp[i-1]没有贡献反而会使dp[i-1]减小，那么就以i又一次作为起点

            {

                s=i;

                t=i;

                dp[i]=a[i];

            }

            else if (a[i]<=dp[i-1]+a[i])  //a[i]比dp[i-1]+a[i]更大，就将尾指针t向后移赋为i

            {

                t=i;

                dp[i]=dp[i-1]+a[i];     //更dp[i]

            }

            if (dp[i]>maxx)   //假设当前dp[i]比之前的最大值要大，更新最大值，并记录首尾指针

            {

                S=s;

                T=t;

                maxx=dp[i];

            }

        }

        pf("Case %d:\n",cas++);

        pf("%d %d %d\n",maxx,S,T);

    }

    return 0;

}

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