最小生成树MST POJ-2485
来源:互联网 发布:腾讯数据分析招聘 编辑:程序博客网 时间:2024/09/21 11:22
Highways
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18191 Accepted: 8455
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
Kruskal算法:
#include<iostream>#include<algorithm>#include<stdio.h>using namespace std;const int MAX = 501;int n,m;int u[MAX*MAX],v[MAX*MAX],w[MAX*MAX],r[MAX*MAX];int p[MAX];int cmp(const int x,const int y) {return w[x]<w[y];}int find(int x) {return (p[x] == x)?x:p[x] = find(p[x]);}//并查集的findint main(){int k,i,j,ans,buf;int num;//选取的边的数目cin>>k;for(;k != 0;k--){cin>>n;m = 0;//边的数目ans = -1;num = 0;for(i = 0;i < n;i++)//读取边for(j = 0;j < n;j++){if(i <= j) {cin>>buf;continue;}//i == j不读u[m] = i;v[m] = j;r[m] = m;cin>>w[m++];}for(i = 0;i < n;i++) p[i] = i;//初始化并查集sort(r,r+m,cmp);for(i = 0;i < m;i++){if(num == n-1) break;//最小生成树n-1条边int e = r[i];int x = find(u[e]);int y = find(v[e]);if(x != y){num++;if(ans < w[e]) ans = w[e];p[x] = y;}}cout<<ans<<endl;}return 0;}
Prim算法:
#include<stdio.h>#include<iostream>using namespace std;const int MAX = 501;#define INF 0x7fffffffint map[MAX][MAX];//邻接矩阵int n;//结点数int lowcost[MAX];//从选中结点集合到相邻边最短距离int vis[MAX];//标志结点有无被选中int main(){int i,j,pos,min,ans,k;scanf("%d",&k);for(;k > 0;k--){scanf("%d",&n);memset(vis,0,sizeof(vis));for(i = 0;i < n;i++)for(j = 0;j < n;j++){scanf("%d",&map[i][j]);}vis[0] = 1; pos = 0;//默认从第一个结点开始for(i = 1;i < n;i++) lowcost[i] = map[pos][i];//初始化最小权值ans = -1;for(j = 1;j < n;j++)//再找出来n-1个点{min = INF;for(i = 1;i < n;i++)if(!vis[i] && min > lowcost[i]){min = lowcost[i];pos = i;}vis[pos] = 1;if(ans < min) ans = min;for(i = 1;i < n;i++)//刷新最小权值if(!vis[i] && lowcost[i] > map[pos][i]){lowcost[i] = map[pos][i];}}printf("%d\n",ans);}return 0;}
- 最小生成树MST POJ-2485
- 最小生成树 MST
- 最小生成树 MST
- MST-最小生成树
- MST 最小生成树
- 最小生成树(MST)
- MST 最小生成树
- poj 1679 The Unique MST (最小生成树)
- Poj-1679 The Unique MST -最小生成树
- POJ 1679 The Unique 次最小生成树 MST
- Poj 1258 Agri-Net[MST最小生成树]
- poj-1679-The Unique MST-最小生成树是否唯一
- poj 1679 The Unique MST(图论:最小生成树)
- poj 1679 The Unique MST 最小生成树
- POJ 1679 The Unique MST【最小生成树问题相关】
- POJ 3026 Borg Maze【BFS+最小生成树MST】
- POJ-1258 Agri-Net 最小生成树(MST)
- poj 1679 The Unique MST (最小生成树是否唯一)
- \tutorial_code\Histograms_Matching
- 囚犯和开关的问题
- 部落卫队问题(分支限界法)
- ID选择器的一些例子
- 大数据的含义
- 最小生成树MST POJ-2485
- SGU 代码集 (Volume 1)
- class类选择器
- void LUT(InputArray src, InputArray lut, OutputArray dst)
- vim+ctags+cscope(工欲善其事,必先利其器。)
- hdu 1068 Girls and Boys (二分图匹配)
- 深入浅出MyBatis-Configuration
- 特殊的选择器
- 如何安装使用SQL