1003-Max Sum

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

 

 

解题思路：

解这道题用的是动态规划,但要考虑的情况却是非常的多，第一重要的也是最容易忽视的就是要考虑当所有的数据都是负数时，这种情况要特殊考虑，第二就是在记录最大和的区间的时候，计算没个f[i]的时候都要将他的区间更新一次。做这题的时候，经过了无数次蛋疼的WA，下面是AC的代码，因为一开始就没写好，很多情况都是后来加上去的，所以代码显得有点繁琐。

 

AC代码：

 #include<iostream>
 #include<stdio.h>
 using namespace std;
 int T,a[1000010];
 struct fun {
  int date;
  int front,rear;
 }f[1000010];
 int ma(int x,int y) {
  return (x>y? x:y);
 }
 int main() {
  scanf("%d",&T);
  for(int l=1;l<=T;++l) {
   int n;
   scanf("%d",&n);
   for(int i=1;i<=n;++i)
   scanf("%d",&a[i]);
   int k=0;
   for(int i=1;i<=n;++i)
   if(a[i]>0)
   k=1;
   int max=1;
   if(k){
    for(int i=0;i<n;++i)
    f[i].front=1;
    f[0].date=0;
    int biaoji=1;
    for(int i=1;i<=n;++i) {
     if(f[i-1].date+a[i]<0 )
      biaoji=i+1;
     else
     f[i].front=biaoji;
     f[i].rear=i;
     f[i].date=ma(0,f[i-1].date+a[i]);
    }
    for(int i=2;i<=n;++i)
    if(f[i].date>f[max].date) {
     max=i;
     f[max].rear=max;
    }
   }
   else if(k==0) {
    int min=1;
    for(int i=2;i<=n;++i)
    if(a[i]>a[min])
    min=i;
    max=min;
    f[max].date=a[min];
    f[max].front=min;
    f[max].rear=min;
   }
   printf("Case %d:\n%d %d %d\n",l,f[max].date,f[max].front,f[max].rear);
   if(l!=T)
   printf("\n");
  }
  return 0;
 }