Description
A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose the location of the fire station so as to reduce the distance to the nearest station from the houses of the disgruntled residents.
The city has up to 500 intersections, connected by road segments of various lengths. No more than 20 road segments intersect at a given intersection. The location of houses and firestations alike are considered to be at intersections (the travel distance from the intersection to the actual building can be discounted). Furthermore, we assume that there is at least one house associated with every intersection. There may be more than one firestation per intersection.
Input
The first line of input contains two positive integers: f,the number of existing fire stations (f <= 100) and i, the number of intersections (i <= 500). The intersections are numbered from 1 to i consecutively. f lines follow; each contains the intersection number at which an existing fire station is found. A number of lines follow, each containing three positive integers: the number of an intersection, the number of a different intersection, and the length of the road segment connecting the intersections. All road segments are two-way (at least as far as fire engines are concerned), and there will exist a route between any pair of intersections.
Output
You are to output a single integer: the lowest intersection number at which a new fire station should be built so as to minimize the maximum distance from any intersection to the nearest fire station.
Sample Input
1 6
2
1 2 10
2 3 10
3 4 10
4 5 10
5 6 10
6 1 10
Sample Output
5
Source
Waterloo local 1999.09.25
题意：已知城市和已建加油站（健在一部分城市），求下一个加油站健在哪里，改城市到加油站最远距离最小
我已无力吐槽这个题了，这个题的坑真是坑死人了，首先那个输入很害人，然后居然用floyd算法，我惊呆了， 看到小伙伴都做出来了，我却wa了11发（第几斯特拉 算法 +误解题意+超时），这个人都不好了， 比赛完看到别人代码，居然还还看不懂，感觉图论太差了(⊙o⊙)…哎。。。。。太菜了
细节我都注释在代码中了（仿 lyf 代码）
#include
#include
#include
#include
using namespace std; #define N 1000 #define maxe 0x3f3f3f3f int a[N][N],dis[N]; int b[N]; int n,m; void inset() { int i,j; for(i=1;i<=m;i++) for(j=1;j<=m;j++) if(i==j) a[i][j]=0; else a[i][j]=a[j][i]=maxe; } void floyd() // 不懂就百度floyd { int i,j,k; for(k=1;k<=m;k++) for(i=1;i<=m;i++) for(j=1;j<=m;j++) if(a[i][j]>a[i][k]+a[k][j]) a[i][j]=a[i][k]+a[k][j]; } void solve() { int i,j,cur=0; int minn; for(i=1;i<=m;i++) { minn=maxe; for(j=0;j
len) a[x][y]=a[y][x]=len; } floyd(); solve(); return 0; }