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  • Conscription poj3723(最大生成树)

    Conscription
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 6870   Accepted: 2361

    Description

    Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

    Input

    The first line of input is the number of test case.
    The first line of each test case contains three integers, NM and R.
    Then R lines followed, each contains three integers xiyi and di.
    There is a blank line before each test case.

    1 ≤ NM ≤ 10000
    0 ≤ R ≤ 50,000
    0 ≤ xi < N
    0 ≤ yi < M
    0 < di < 10000

    Output

    For each test case output the answer in a single line.

    Sample Input

    2
    
    5 5 8
    4 3 6831
    1 3 4583
    0 0 6592
    0 1 3063
    3 3 4975
    1 3 2049
    4 2 2104
    2 2 781
    
    5 5 10
    2 4 9820
    3 2 6236
    3 1 8864
    2 4 8326
    2 0 5156
    2 0 1463
    4 1 2439
    0 4 4373
    3 4 8889
    2 4 3133
    

    Sample Output

    71071
    54223

     1 #include <iostream>
     2 #include <string.h>
     3 #include <stdio.h>
     4 #include <algorithm>
     5 using namespace std;
     6 typedef struct abcd
     7 {
     8     int x,y,z;
     9 } abcd;
    10 bool cmp(abcd x,abcd y)
    11 {
    12     return x.z>y.z;
    13 }
    14 abcd a[70000];
    15 int p[100000];
    16 int findset(int x)
    17 {
    18     int i,px=x;
    19     while(px!=p[px])px=p[px];
    20     while(x!=px)
    21     {
    22         i=p[x];
    23         p[x]=px;
    24         x=i;
    25     }
    26     return px;
    27 }
    28 int main()
    29 {
    30     int t,n,m,r,i,j,x,y,z;
    31     cin>>t;
    32     while(t--)
    33     {
    34         scanf("%d%d%d",&n,&m,&r);
    35         for(i=0; i<m+n; i++)p[i]=i;
    36         for(i=0; i<r; i++)
    37         {
    38             scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
    39             a[i].y+=n;
    40         }
    41         sort(a,a+r,cmp);
    42         long long sum=10000*(m+n);
    43         for(i=0; i<r; i++)
    44         {
    45             int x1=findset(a[i].x),y1=findset(a[i].y);
    46             if(x1!=y1)
    47             {
    48                 sum-=a[i].z;
    49                 p[x1]=y1;
    50             }
    51         }
    52         cout<<sum<<endl;
    53     }
    54 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3589547.html
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