【POJ】-2485-Highways(最小生成树)
来源:互联网 发布:怎么解决淘宝自主访问 编辑:程序博客网 时间:2024/05/17 03:13
Highways
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28432 Accepted: 12958
Description
The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible to drive between any pair of towns without leaving the highway system.
Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.
The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.
Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.
The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.
Input
The first line of input is an integer T, which tells how many test cases followed.
The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.
The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.
Output
For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.
Sample Input
130 990 692990 0 179692 179 0
Sample Output
692
矩阵形式的表示边的权值的题目,注意输出什么,读题!
#include<cstdio>#include<cstring>#include<queue>#include<stack>#include<cmath>#include<vector>#include<algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))struct node{int from,to,va;}a[1100000];bool cmp(node x,node y){return x.va<y.va;}int f[510];int find(int x){if(x!=f[x])f[x]=find(f[x]);return f[x];}bool join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){f[fx]=fy;return true;} return false;}int main(){int u;scanf("%d",&u);while(u--){int n;scanf("%d",&n);for(int i=1;i<=n;i++)f[i]=i;int t,num=1;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(j>i)//只看一半,减少时间 {scanf("%d",&t);a[num].from=i;a[num].to=j;a[num].va=t;num++;}elsescanf("%d",&t);}}sort(a+1,a+num,cmp);int ant=0;num=0;for(int i=1;i<=n*(n-1)/2;i++){if(join(a[i].from,a[i].to)){ant=max(ant,a[i].va);//输出所有连接的边里最长的 num++;}if(num==n-1)break;}printf("%d\n",ant);}return 0;}
0 0
- POJ 2485 Highways (最小生成树)
- poj 2485 Highways (最小生成树)
- POJ 2485 Highways(最小生成树)
- 【POJ】-2485-Highways(最小生成树)
- POJ 2485 Highways(最小生成树)
- POJ 2485Highways(最小生成树)
- 【POJ 2485】Highways(最小生成树)
- POJ 2485 Highways【最小生成树】
- poj 2485Highways(最小生成树 Kruskal)
- poj 2485 Highways (最小生成树)
- poj 2485 Highways 最小生成树
- poj 2485 Highways prim最小生成树
- POJ 2485 Highways 最小生成树
- POJ 2485 Highways(最小生成树)
- POJ 2485 Highways 最小生成树
- poj 2485 Highways 最小生成树
- POJ 2485 Highways (最小生成树)
- POJ 2485 Highways (prim最小生成树)
- JAVA集合容器----TreeMap、TreeSet
- Linux动态链接(2)so初始化执行
- 133.Your database is open and the LISTENER listener is running. The new DBA of the system stops the
- Android style样式的简单理解
- 对员工的增加,删除
- 【POJ】-2485-Highways(最小生成树)
- python TCP编程小试牛刀
- 深入理解 Java 注解 [元注解(一)]
- 工作流现状2008年
- 堆排序
- #5 Deep learning RNN-RBM简单理解
- Android图片加载与缓存开源框架:Android Glide
- Struts2 我的第一个Helloworld
- Rxlifecycle使用详解,解决RxJava内存泄露问题