POJ 2607 Fire Station（floyd）

POJ 2607 Fire Station（floyd）

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

題意：已知城市和已建加油站（健在一部分城市），求下一個加油站健在哪裡，改城市到加油站最遠距離最小


我已無力吐槽這個題了，這個題的坑真是坑死人了，首先那個輸入很害人，然後居然用floyd算法，我驚呆了， 看到小伙伴都做出來了，我卻wa了11發（第幾斯特拉 算法 +誤解題意+超時），這個人都不好了， 比賽完看到別人代碼，居然還還看不懂，感覺圖論太差了(⊙o⊙)…哎。。。。。太菜了
細節我都注釋在代碼中了（仿 lyf 代碼）

#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;jlen)
          a[x][y]=a[y][x]=len;
    }

    floyd();

    solve();

    return 0;
}