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=;
     while(test--)
     {
         cin>>n;
         for(i=;i<=n;i++)
             cin>>a[i];
         d[]=a[];
         for(i=;i<=n;i++)
         {
             if(d[i-]<) d[i]=a[i];
             else d[i]=d[i-]+a[i];
         }
         max=d[];e=;
         for(i=;i<=n;i++)
         {
             if(max<d[i])
             {
                 max=d[i];e=i;
             }
         }
         int t=;
         f=e;
         for(i=e;i>;i--)
         {
             t=t+a[i];
             if(t==max)    f=i;
         }
         cout<<"Case "<<k++<<":"<<endl<<max<<" "<<f<<" "<<e<<endl;
         if(test) cout<<endl;
     }
     return ;
 }

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=;i<n;i++)
         {
             scanf("%d",&a);
             if(t<) t=a;
             else t=t+a;
             if(ans<t) ans=t;
         }
         printf("%d\n",ans);
     }
     return ;
 }

 #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 = ;

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

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

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

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

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 = ;

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

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

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

             if (sum <= ) {//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:\n", ++cas);
         printf("%d %d %d\n", ans, bg + , ed + );
         if (T > ) {
             printf("\n");
         }
     }
     return ;
 }