zoukankan      html  css  js  c++  java
  • poj 2377 Bad Cowtractors(最大生成树!)

    Description

    Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie. 
    
    Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

    Input

    * Line 1: Two space-separated integers: N and M 
    
    * Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

    Output

     

    * Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

     

     

    Sample Input
    5 8
    1 2 3
    1 3 7
    2 3 10
    2 4 4
    2 5 8
    3 4 6
    3 5 2
    4 5 17

    Sample Output

    42

    Hint

    OUTPUT DETAILS: 
    
    The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

    Source

     
    其实不过就是裸的最小生成树,(kruscal实现),只不过这里的话要将边按从大到小排序,然后判断一下是不是连通的就可以了。
     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<math.h>
     7 #include<algorithm>
     8 #include<queue>
     9 #include<set>
    10 #include<bitset>
    11 #include<map>
    12 #include<vector>
    13 #include<stdlib.h>
    14 #include <stack>
    15 using namespace std;
    16 #define PI acos(-1.0)
    17 #define max(a,b) (a) > (b) ? (a) : (b)  
    18 #define min(a,b) (a) < (b) ? (a) : (b)
    19 #define ll long long
    20 #define eps 1e-10
    21 #define MOD 1000000007
    22 #define N 1006
    23 #define M 20006
    24 #define inf 1e12
    25 struct Node{
    26     int x,y;
    27     int cost;
    28 }edge[M];
    29 int n,m;
    30 int fa[N];
    31 void init(){
    32     for(int i=0;i<N;i++){
    33         fa[i]=i;
    34     }
    35 }
    36 int find(int x){
    37     return fa[x]==x?x:fa[x]=find(fa[x]);
    38 }
    39 bool cmp(Node a,Node b){
    40     return a.cost>b.cost;
    41 }
    42 int main()
    43 {
    44     while(scanf("%d%d",&n,&m)==2){
    45         init();
    46         for(int i=0;i<m;i++){
    47             int a,b,c;
    48             scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
    49         }
    50         sort(edge,edge+m,cmp);
    51         int ans=0;
    52         int num=n-1;
    53         for(int i=0;i<m;i++){
    54             int root1=find(edge[i].x);
    55             int root2=find(edge[i].y);
    56             if(root1!=root2){
    57                 ans+=edge[i].cost;
    58                 fa[root1]=root2;
    59                 num--;
    60             }
    61         }
    62         if(num!=0){
    63             printf("-1
    ");
    64         }
    65         else{
    66             printf("%d
    ",ans);
    67         }
    68     }
    69     return 0;
    70 }
    View Code
  • 相关阅读:
    OpenCv 109---Blob分析
    OpenCv 013---图像的翻转
    OpenCv 012---视频文件读写
    MenuExample
    OpenCv 011---像素归一化
    OpenCv 010---图像像素值统计
    OpenCv 009---色彩空间与色彩空间转换
    OpenCv 008---通道分离与合并
    OpenCv 007---像素操作的逻辑操作
    OpenCv 006---LUT的作用与用法
  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4817725.html
Copyright © 2011-2022 走看看