NOIP2014 寻找道路

传送门

这道题还是比较简单的。我们只要先用老套路建出反图，记录终点与哪些点是联通的，之后从所有不与终点联通的点出发，在反图上枚举一下与之直接相连的边，也设为不能走。之后我们在可以走的路上跑最短路即可。

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<set>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define sc second
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = 100005;
const int INF = 1000000009;

int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') ans *= 10,ans += ch - '0',ch = getchar();
    return ans * op;
}

struct edge
{
    int next,to,from;
}e[M<<2],e1[M<<2];

int n,m,head[M],ecnt,dis[M],S,T,head1[M],ecnt1,x,y;
bool vis[M],pd[M];
queue <int> Q;
set <pr> q;
set <pr> :: iterator it;

void add(int x,int y)
{
    e[++ecnt].to = y;
    e[ecnt].next = head[x];
    e[ecnt].from = x;
    head[x] = ecnt;
}

void add1(int x,int y)
{
    e1[++ecnt1].to = y;
    e1[ecnt1].next = head1[x];
    e1[ecnt1].from = x;
    head1[x] = ecnt1;
}

void bfs()
{
    Q.push(T);vis[T] = 1;
    while(!Q.empty())
    {
        int k = Q.front();Q.pop();
        for(int i = head1[k];i;i = e1[i].next)
        {
            int v = e1[i].to;
            if(!vis[v]) vis[v] = 1,Q.push(v);
        }
    }
}

void dij()
{
    dis[S] = 0;
    q.insert(mp(0,S));
    while(!q.empty())
    {
        pr k = *(q.begin());q.erase(q.begin());
        for(int i = head[k.sc];i;i = e[i].next)
        {
            if(!vis[e[i].to] || pd[e[i].to]) continue;
            if(dis[e[i].to] > dis[k.sc] + 1) 
            {
                it = q.find(mp(dis[e[i].to],e[i].to));
                if(it != q.end()) q.erase(it);
                dis[e[i].to] = dis[k.sc] + 1;
                q.insert(mp(dis[e[i].to],e[i].to));
            }
        }
    }
}

int main()
{
    n = read(),m = read();
    rep(i,1,n) dis[i] = INF;
    rep(i,1,m) x = read(),y = read(),add(x,y),add1(y,x);
    S = read(),T = read();
    bfs();
    rep(i,1,n) 
    if(!vis[i])
    {
         for(int j = head1[i];j;j = e1[j].next) pd[e1[j].to] = 1;
    }
    dij();
    if(dis[T] == INF) printf("-1\n");
    else printf("%d\n",dis[T]);
    return 0;
}