POJ 2485 Highways中如何设计高速公路实现最短路径计算？

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题目：道路时限：1000MS 内存限制：65536KB 64位IO格式：%I64d %I64u 提交状态：练习 POJ 2485 系统爬虫（2014-09-09） 描述：岛屿国家Flatopia地形平坦。不幸的是，Flatopia没有公共高速公路。

Highways

Time Limit:1000MSMemory Limit:65536KB64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2485

System Crawler (2014-09-09)

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.

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.

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

1 3 0 990 692 990 0 179 692 179 0

Sample Output

692

Hint

Huge input,scanf is recommended.

#include<stdio.h> #include<string.h> #include<stdlib.h> #define N 1005 #define INF 0x7fffffff int map[N][N],v[N]; int num[N]; int n; void prime() { int s = 0; int i,j; int max = 0; memset(v,0,sizeof(v)); for(i=0;i<=n;i++) { num[i] = INF; } for(i=0;i<n;i++) { num[i] = map[s][i]; } num[s] = 0; for(i=0;i<n;i++) { int min = INF,k; for(j=0;j<n;j++) { if(v[j] == 0 && num[j]<min) { min = num[j]; k = j; } } if(min == INF) { break; } if(max<min) { max = min; } v[k] = 1; for(j=0;j<n;j++) { if(v[j] == 0 && num[j]>map[j][k]) { num[j] = map[j][k]; } } } printf("%d\n",max); } int main() { int i,j; int T; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=0;i<n;i++) { for(j=0;j<n;j++) { scanf("%d",&map[i][j]); } } prime(); } return 0; }