• 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
    ",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:
    ", ++cas);
    40         printf("%d %d %d
    ", ans, bg + 1, ed + 1);
    41         if (T > 0) {
    42             printf("
    ");
    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:
    ", ++cas);
    44         printf("%d %d %d
    ", ans, bg + 1, ed + 1);
    45         if (T > 0) {
    46             printf("
    ");
    47         }
    48     }
    49     return 0;
    50 }
    View Code
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  • 原文地址:https://www.cnblogs.com/gongpixin/p/6735454.html
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