POJ
来源:互联网 发布:mac os x10.7.5升级包 编辑:程序博客网 时间:2024/06/14 08:45
问题描述
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 81 2 31 3 72 3 102 4 42 5 83 4 63 5 24 5 17
Sample Output
42
分析:
代码如下:
#include<cstdio>#include<algorithm>using namespace std;const int maxn = 1000+10;const int maxe = 20000+10;struct edge{ int u, v, cost;}es[maxe];int cmp(const edge &a, const edge &b){ return a.cost > b.cost;}int N,M;int par[maxn];/*初始化*/void Init(){ for(int i=1; i<=N; i++) { par[i] = i; }}/*找爹*/int find(int x){ int r = x; while(r!=par[r]) r = par[r]; int i = x; int j; while(i != r) { j = par[i]; par[i] = r; i = j; } return r;}/*是不是同个爹*/bool same(int x, int y){ return find(x)==find(y);}/*合并爹*/void unite(int x, int y){ par[find(x)] = find(y);}/*最小生成树算法*/int kruskal(){ sort(es, es+M, cmp); Init(); int res = 0; for(int i=0; i<M; i++) { edge e = es[i]; if(!same(e.u, e.v)) { res += e.cost; unite(e.u, e.v); } } return res;}int main(){ scanf("%d%d",&N,&M); for(int i=0; i<M; i++) { scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].cost); } int k = 0; kruskal(); for(int i=1; i<=N; i++) { if(par[i] == i) k++; } if(k>1) printf("-1\n"); else printf("%d\n", kruskal()); return 0;}
- POJ
- poj
- POJ
- POJ
- poj
- poj
- POJ
- POJ
- poj
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- UVa 232 Crossword Answers
- shell
- opencv 2D直方图
- ListView+ SQLite实现商品展示
- Android:AS与Unity3D之间打包的各种坑及解决方案
- POJ
- IF脚本实现虚拟机的开启,关闭,重置,快照
- Hadoop2.x 让你真正明白yarn
- Android Arm Inline Hook
- Android出现“Read-only file system”解决办法
- Adaboost原理简析
- mac install kafka
- 数据库基础
- poj 3164 Command Network(有定根的最小树形图)