poj-2485 Highways


Highways Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 30003 Accepted: 13656 Descrtion

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 oblem. 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.

男人的天堂 AV电影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.

男人的天堂 AV电影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


3 0 990 692 990 0 179 692 179 0

Sample Output



Huge input,scanf is recommended.

描述 :有個城市叫做H市。其中有很多個村莊,村莊之間通信基本靠吼,交通基本靠走,很不方便。 這個市長知道了這個情況,為了替市民著想,決定修建高鐵。每修建一米花費1美元。 現在市長請了最著名的工程師來修建高鐵,自然這個工程師會讓修建高鐵的費用最少。 不幸的是,在修建了高鐵之后就病逝了。現在市長希望知道在修建完成的這些高鐵路中最長的一段高鐵路花費了多少美元, 他請你來幫助他,如果你計算正確,市長將會送你一輛蘭博基尼。 輸入: 第一行一個數T,表示接下來有多少組數據。 接下來每組測試數據的第一行有一個數N(3<=N<=500), 表示村莊數目。 然后是一個二維數組,第 i行第j列表示第i個村莊到第j個村莊的距離。 輸出: 只有一個數,輸出市長希望知道的已經修成的高鐵中最 長的路花了多少錢。


#include <stdio.h> #define min(a,b) a<b?a:b #define max(a,b) a>b?a:b int map[505][505]; int vis[505],s[505];// vis表示該點到每個頂點的最小值,s標記頂點 int inf=999999; int main() { int t,n; scanf("%d",&t); while(t--) { int i,j; int res[505],ans; scanf("%d",&n); for(i=1;i<=n;i++){ vis[i]=inf; s[i]=0;// } vis[1]=0; for(i=1;i<=n;i++){ for(j=1;j<=n;j++){ scanf("%d",&map[i][j]); } } int f=1; for(int k=1;k<=n;k++){ int v=0; for(i=1;i<=n;i++){ if(!s[i] && (v==0||vis[i]<vis[v])) v=i; } if(i==0) break; s[v]=1;//標記 res[f++]=vis[v]; // res+=vis[v]; for(j=1;j<=n;j++){//松弛操作 if(!s[i]) vis[j]=min(vis[j],map[v][j]); } } for(i=2;i<=n;i++){ ans=max(res[i-1],res[i]); } printf("%d\n",ans); } return 0; }


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