链接：
Horror List
It was time for the 7th Nordic Cinema Popcorn Convention, and this year the manager Ian had a brilliant idea. In addition to the traditional film program, there would be a surprise room where a small group of people could stream a random movie from a large collection, while enjoying popcorn and martinis.
However, it turned out that some people were extremely disappointed, because they got to see movies like Ghosts of Mars, which instead caused them to tear out their hair in despair and horror.
To avoid this problem for the next convention, Ian has come up with a solution, but he needs your help to implement it. When the group enters the surprise room, they will type in a list of movies in a computer. This is the so-called horror list, which consists of bad movies that no one in the group would ever like to see. Of course, this list varies from group to group.
You also have access to the database Awesome Comparison of Movieswhich tells you which movies are directly similar to which. You can assume that movies that are similar to bad movies will be almost as bad. More specificly, we define the Horror index as follows:
Input
The first line of input contains three positive integers N , H , L ( 1≤H<N≤1000,0≤L≤10000 ), where N is the number of movies (represented by IDs, ranging from 0 to N−1 ), H is the number of movies on the horror list and L is the number of similarities in the database.
The second line contains H unique space-separated integers xi ( 0≤xi < N ) denoting the ID of the movies on the horror list.
The following L lines contains two space-separated integers ai , b i ( 0≤ai < b i < N ), denoting that movie with ID ai is similar to movie with ID bi (and vice verca).
Output
Output the ID of the best movie in the collection (highest Horror Index). In case of a tie, output the movie with the lowest ID.
Sample Input
6 3 5 0 5 2 0 1 1 2 4 5 3 5 0 2
Sample Output
1
Source
算法：最短路 =_= !
题意+思路：
根据上图中的公式，求出每个电影的恐怖值。。。
PS：最终你会发现，最恐怖的电影的价值应该是 0， 最不恐怖的反而尽量大。。。这一点很容易搞反。
第一行 HI = 0 ：表示在恐怖表中的电影，恐怖值均为 0 ，而且不会被其他的关系定义覆盖。【相当于以这些电影为起点】
第二行 HI = Q+1：如果和当前电影类似的最恐怖的电影的恐怖值是 Q ，那么当前电影的 HI = Q+1
【相当于起点到当前电影的最小距离】
第三行 HI = INF ：如果当前电影不和任何电影相似。【表示是孤立的点,最不恐怖的电影】
一般的大多是正常人，为了不把自己吓的抓狂，当然是选择不怎么恐怖的电影看了，这里为了尽量满足大多数人的需求，当然是选择最不恐怖的电影了。如果恐怖度一样，就选择编号最小的电影。
那么题目就转化成了求 HI 最大的，电影编号，如果有多个一样，则输出最小的。PS：如果有INF的，当然直接输出最小的编号就over了
相当于以恐怖名单上的电影编号为起点，d[index] = 0; 如果 index 出现在恐怖名单上【样例中的第二行】
然后每一个点【电影】为终点，求出起点集合到终点的最短距离。
最后输出最短距离中距离最大的那个。。。
不知道说清楚了没有，反正比赛的时候我是没有想明白的，完了听Orc 提别人说是最短路才反应过来。。。
code：
//PS:用邻接矩阵写的一个比较丑的代码,还是不习惯用邻接表写。。。等熟悉了再折腾吧
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxnN = 1000+10;
const int maxnL = 10000+10;
const int INF = 9999999;
int n,h,L;
int d[maxnN];
int w[maxnN][maxnN];
int vis[maxnN];
void Dijkstra()
{
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++)
{
int x , m = INF;
for(int y = 0; y < n; y++) if(!vis[y] && d[y] <= m) m = d[x=y];
if(m == INF) return; //如果存在孤立的点,不用继续判断
vis[x] = 1;
for(int y = 0; y < n; y++)
d[y] = min(d[y], d[x]+w[x][y]);
}
}
int main()
{
while(scanf("%d%d%d", &n,&h,&L) != EOF)
{
for(int i = 0; i < n; i++) d[i] = INF; //初始化每一条边,都没有相邻的
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
w[i][j] = (i == j ? 0 : INF);
int x;
for(int i = 0; i < h; i++)
{
scanf("%d", &x);
d[x] = 0; //在恐怖表中的电影
}
int a,b;
for(int i = 0; i < L; i++)
{
scanf("%d%d", &a,&b);
w[a][b] = 1; //双向图
w[b][a] = 1;
}
Dijkstra();
int Max = d[0]; //从第一个开始查找
int index = 0;
for(int i = 1; i < n; i++)
{
if(d[i] > Max)
{
Max = d[i];
index = i;
}
}
printf("%d
", index);
//for(int i = 0; i < n; i++) printf("%d ", d[i]); printf("
");
}
return 0;
}