URAL 2034 : Caravans

Description

Student Ilya often skips his classes at the university. His friends criticize him for this, but they don’t know that Ilya spends this time not watching TV serials or listening to music. He creates a computer game of his dreams. The game’s world is a forest. There are elves, wooden houses, and a villain. And one can rob caravans there! Though there is only one caravan in the game so far, Ilya has hard time trying to implement the process of robbing.

The game world can be represented as several settlements connected by roads. It is possible to get from any settlement to any other by roads (possibly, passing through other settlements on the way). The settlements are numbered by integers from 1 to n. All the roads are two-way and have the same length equal to 1. It is not allowed to move outside the roads. The caravan goes from settlement s to settlement f following one of the shortest routes. Settlement r is the villain’s den. A band of robbers from settlement r has received an assignment to rob the caravan. In the evening they will have an exact plan of the route that the caravan will take the next morning. During the night they will be able to move to any settlement on the route, even to settlement s or f. They will lay an ambush there and rob the caravan in the morning. Of course, among all such settlements the robbers will choose the one closest to settlement r. The robbers have a lot of time until the evening. They don’t know the caravan’s route yet, but they want to know the maximum distance they will have to go in the worst case to the settlement where they will rob the caravan. Help Ilya calculate this distance, and it may happen that he will attend his classes again!

Input

The first line contains integers n and m (3 ≤ n ≤ 10 5; 2 ≤ m ≤ 10 5), which are the number of settlements in the game world and the number of roads between them. Each of the following m lines describes a road. The description contains integers a and b, which are the numbers of the settlements connected by the road. It is guaranteed that each road connects two different settlements and there is at most one road between any two settlements. It is also guaranteed that the road network is connected. In the last line you are given pairwise different integers s, f, and r, which are the numbers of the settlements described above.

Output

In the only line output the required distance.

input	output
7 7 1 2 2 4 2 5 3 4 4 6 5 6 6 7 1 7 3	2

题目大意：
大意是有一支商队要从点 s 到点 f,这个商队只会走走短路（题目保证能从 s 到 f ，同时，走每条边的时间花费都是 1），有一个强盗在点 r ,这个强盗要去抢商队
就问你在最坏情况下，强盗能够拦截到这个商队所需走的最短距离.
比如样例：



这里商队需要从点 1 到点 7 ，强盗在点 3，我们可以很容易知道商队的路线有两条:
1 -> 2 -> 4 ->6 -> 7
1 -> 2 -> 5 ->6 -> 7
而在最坏情况下，也就是商队走1->2->5->6->7时，强盗在点 3 想要拦截到商队需要至少走两步（到 2 或者 6 ） 才能拦截到商队
所以样例的输出是 2

解题报告：
首先我们肯定需要求一次最短路，这个最短路是强盗所在的点 r ，到其他各个点的最短路，我们不妨设置值为 maxv[]（不要奇怪这个命名）
即maxv[ x ]表示点 r 到点 x 的最短距离.
显然我们这道题需要使用到DP（递推）求解，注意到这些递推的点必须是最短路上的点同时也必须只能转移到最短路上的点，那么我们该如何确定这个最短路的点呢？
我们令 d[] 这个数组表示从终点 f 到其他各个点的最短距离，那么我们只要在转移时必须满足 d[v] == d[u] - 1 即可(从u转移到v) <仔细想想>
这样我们就保证了递推过程的点肯定是最短路上的点，也只能转移到最短路上的点
之后我们考虑DP值,我们令DP[ i ] 表示拦截从 起点 到 i 这个点所在的最短路的最小花费
那么我们很容易的就可以得出递推方程（用 u 去更新 v)
dp[v] = max( maxv[u] } <仔细想想> ，取max是因为我们要保证最差情况

那么当我们扩展到点 u时，需要首先更新 maxv 值，即 maxv[u] = min{ maxv[u] , dp[u] } ，这个方程就非常容易了，因为拦截从 起点 到 点i所在的最短路不外乎就两种方式，一种在前面拦截，第二种在这个点 i 拦截，所以我们取个min即可(强盗也要走最短的对不对)

那么，我们就解决了这道题

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 1e5 + 50;
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
vector<int>e[maxn];
int dp[maxn]  , d[maxn] , maxv[maxn], n , m , s , f , r , used[maxn] ;
queue<int>q;

void solve()
{
memset(dp , 0 , sizeof(dp));memset(d,-1,sizeof(d));memset(maxv,-1,sizeof(maxv));memset(used,0,sizeof(used));
d[f] = 0;
q.push(f);
while(!q.empty())
{
int x = q.front();q.pop();
for(int i = 0 ; i < e[x].size() ; ++ i)
{
int v = e[x][i];
if (d[v] == -1)
{
d[v] = d[x] + 1;
q.push(v);
}
}
}
maxv[r] = 0;
q.push(r);
while(!q.empty())
{
int x = q.front();q.pop();
for(int i = 0 ; i < e[x].size() ; ++ i)
{
int v = e[x][i];
if (maxv[v] == -1)
{
maxv[v] = maxv[x] + 1;
q.push(v);
}
}
}
dp[s] = maxv[s];used[s] = 1;
q.push(s);
while(!q.empty())
{
int x = q.front();q.pop();
maxv[x] = min(maxv[x],dp[x]);
for(int i = 0 ; i < e[x].size() ; ++ i)
{
int v = e[x][i];
if (d[x] - d[v] == 1)
{
dp[v] = max(dp[v],maxv[x]);
if (!used[v])
{
q.push(v);
used[v] = 1;
}
}
}
}
printf("%d\n",maxv[f]);
}

int main(int argc,char *argv[])
{
scanf("%d%d",&n,&m);
while(m--)
{
int u ,v ;
scanf("%d%d",&u,&v);u--,v--;
e[u].push_back(v);e[v].push_back(u);
}
scanf("%d%d%d",&s,&f,&r);s--,f--,r--;
solve();
return 0;
}