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} ;
 1 #include<iostream>
 2 #define N 100010
 3 using namespace std;
 4 int a[N],d[N];
 5 int main()
 6 {
 7     int test,n,i,max,k,f,e;
 8     cin>>test;
 9     k=1;
10     while(test--)
11     {
12         cin>>n;
13         for(i=1;i<=n;i++)
14             cin>>a[i];
15         d[1]=a[1];
16         for(i=2;i<=n;i++)
17         {
18             if(d[i-1]<0) d[i]=a[i];
19             else d[i]=d[i-1]+a[i];
20         }
21         max=d[1];e=1;
22         for(i=2;i<=n;i++)
23         {
24             if(max<d[i])
25             {
26                 max=d[i];e=i;
27             }
28         }
29         int t=0;
30         f=e;
31         for(i=e;i>0;i--)
32         {
33             t=t+a[i];
34             if(t==max)    f=i;
35         }
36         cout<<"Case "<<k++<<":"<<endl<<max<<" "<<f<<" "<<e<<endl;
37         if(test) cout<<endl;
38     }
39     return 0;
40 }
View Code

改进后的只处理最大和不处理位置

 1 #include<cstdio>
 2 int main()
 3 {
 4     int n,test,ans,t,a,i;
 5     scanf("%d",&test);
 6     while(test--)
 7     {
 8         scanf("%d",&n);
 9         scanf("%d",&a);
10         ans=t=a;
11         for(i=1;i<n;i++) 
12         {
13             scanf("%d",&a);
14             if(t<0) t=a;
15             else t=t+a;
16             if(ans<t) ans=t;
17         }
18         printf("%d\n",ans);
19     }
20     return 0;
21 }
View Code

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int T;
 7     scanf("%d", &T);
 8 
 9     int N;
10     int a;
11     int ans;
12     int sum;
13     int i;
14     int bg, ed;//起始,结束
15     int bg2;
16     int cas = 0;
17 
18     while (T--) {
19         scanf("%d", &N);
20 
21         ans = -1010;//
22         sum = 0;//
23         bg2 = 0;//默认起始位置
24         for (i = 0; i < N; ++i) {
25             scanf("%d", &a);
26 
27             sum = sum + a;
28             if (sum > ans) {
29                 ans = sum;
30                 bg = bg2;
31                 ed = i;
32             }
33             if (sum < 0) {//< 0 那么这段到此为止吧
34                 sum = 0;//
35                 bg2 = i + 1;//更新起始位置
36             }
37         }
38 
39         printf("Case %d:\n", ++cas);
40         printf("%d %d %d\n", ans, bg + 1, ed + 1);
41         if (T > 0) {
42             printf("\n");
43         }
44     }
45     return 0;
46 }
View Code

这个和上面的相比写法容易理解些

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int T;
 7     scanf("%d", &T);
 8 
 9     int N;
10     int a;
11     int ans;
12     int sum;
13     int i;
14     int bg, ed;//起始,结束
15     int bg2;
16     int cas = 0;
17 
18     while (T--) {
19         scanf("%d", &N);
20 
21         scanf("%d", &a);
22         ans = a;
23         sum = a;
24         bg = 0, ed = 0;
25         bg2 = 0;
26 
27         for (i = 1; i < N; ++i) {
28             scanf("%d", &a);
29 
30             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
31                 sum = a;
32                 bg2 = i;
33             } else {
34                 sum = sum + a;
35             }
36 
37             if (sum >= ans) {//这里> 和>= 都可以
38                 ans = sum;
39                 bg = bg2, ed = i;
40             }
41         }
42 
43         printf("Case %d:\n", ++cas);
44         printf("%d %d %d\n", ans, bg + 1, ed + 1);
45         if (T > 0) {
46             printf("\n");
47         }
48     }
49     return 0;
50 }
View Code

 

posted @ 2017-04-19 20:18  gongpixin  阅读(263)  评论(0编辑  收藏  举报