題干：

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

13 0 990 692 990 0 179 692 179 0

Sample Output

692

Hint

Huge input,scanf is recommended.

題目大意：

A國沒有高速公路，因此A國的交通很困難。政府意識到了這個問題并且計劃建造一些高速公路，以至于可以在不離開高速公路的情況下在任意兩座城鎮之間行駛。

A國的城鎮編號為1到N, 每條高速公路連接這兩個城鎮，所有高速公路都可以在兩個方向上使用。高速公路可以自由的相互交叉。

A國政府希望盡量減少最長高速公路的建設時間（使建設的最長的高速公路最短），但是他們要保證每個城鎮都可以通過高速公路到達任意一座城鎮。

解題報告：

? ?介于有的題目確實Prim算法效率極高（對于n=2000這樣的完全圖，用Kruskal就作死了），我決定順道學習一下Prim、、、（

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; const int INF = 0x3f3f3f3f; int n; int cost[666][666]; int dis[666]; bool vis[666]; int prim() {//這種寫法適用于已知點是1~N的。int res = 0;for(int i = 0; i<=n; i++) dis[i] = INF, vis[i] = 0;dis[1] = 0;//假設從1開始while(1) {int v = 0;//這種寫法的話上面初始化必須從0開始了。。for(int i = 1; i<=n; i++) {if(!vis[i] && dis[i] < dis[v]) v = i;}if(!v) break;//這種寫法的話上面初始化必須從0開始了。。 vis[v] = 1;res = max(res,dis[v]) ;for(int i = 1; i<=n; i++) {if(!vis[i] && cost[v][i] < dis[i]) dis[i] = cost[v][i];//!vis[i]這一句加不加都一樣的。}} return res; } int main() {int t;cin>>t;while(t--) {scanf("%d",&n);for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {scanf("%d",&cost[i][j]);} }int ans = prim();printf("%d\n",ans);}return 0 ;}

Prim的另一種寫法：（相對常用一些）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; const int INF = 0x3f3f3f3f; int n; int cost[666][666]; int dis[666]; bool vis[666]; int prim() {//這種寫法不一定必須點是1~N的，也可以是0~N-1的，只需要把下面第三行那個改成dis[0]=0就行了。int res = 0;for(int i = 1; i<=n; i++) dis[i] = INF, vis[i] = 0;//這種寫法的話初始化就可以從1開始 dis[1] = 0;//假設從1開始while(1) {int v,minw = INF;for(int i = 1; i<=n; i++) {if(!vis[i] && dis[i] < minw) v = i,minw = dis[i];}if(minw == INF) break;vis[v] = 1;res = max(res,dis[v]) ;for(int i = 1; i<=n; i++) {if(!vis[i] && cost[v][i] < dis[i]) dis[i] = cost[v][i];}} return res; } int main() {int t;cin>>t;while(t--) {scanf("%d",&n);for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {scanf("%d",&cost[i][j]);} }int ans = prim();printf("%d\n",ans);}return 0 ;}

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