NOIP2014D2T2寻找道路

洛谷传送门

这道题可以把边都反着存一遍，从终点开始深搜，然后把到不了的点 和它们所指向的点都去掉。

最后在剩余的点里跑一遍spfa就可以了。

——代码

#include <cstdio>
#include <cstring>
#include <queue>

const int maxn = ;
int n, m, s, t, cnt1, cnt2;
int dis[maxn];
int next1[ * maxn], to1[ * maxn], head1[ * maxn],
    next2[ * maxn], to2[ * maxn], head2[ * maxn];
bool vis[maxn], b[maxn], f[maxn], flag;

inline void add1(int x, int y)
{
    to1[cnt1] = y;
    next1[cnt1] = head1[x];
    head1[x] = cnt1++;
}

inline void add2(int x, int y)
{
    to2[cnt2] = y;
    next2[cnt2] = head2[x];
    head2[x] = cnt2++;
}

inline void dfs(int u)
{
    int i, v;
    if(u == s) flag = ;
    vis[u] = ;
    for(i = head2[u]; i != -; i = next2[i])
    {
        v = to2[i];
        if(!vis[v]) dfs(v);
    }
}

inline void spfa(int u)
{
    int i, j;
    std::queue <int> q;
    q.push(u);
    f[u] = ;
    dis[u] = ;
    while(!q.empty())
    {
        i = q.front();
        q.pop();
        f[i] = ;
        for(j = head1[i]; j != -; j = next1[j])
         if(!b[i] && !b[to1[j]] && dis[to1[j]] > dis[i] + )
         {
             dis[to1[j]] = dis[i] + ;
             if(!f[to1[j]])
             {
                 f[to1[j]] = ;
                 q.push(to1[j]);
             }
         }
    }
}

int main()
{
    int i, j, x, y;
    memset(head1, -, sizeof(head1));
    memset(head2, -, sizeof(head2));
    memset(dis,  / , sizeof(dis));
    scanf("%d %d", &n, &m);
    for(i = ; i <= m; i++)
    {
        scanf("%d %d", &x, &y);
        add1(x, y);
        add2(y, x);
    }
    scanf("%d %d", &s, &t);
    dfs(t);
    if(!flag)
    {
        printf("-1");
        return ;
    }
    for(i = ; i <= n; i++)
     if(!vis[i])
     {
         b[i] = ;
         for(j = head2[i]; j != -; j = next2[j]) b[to2[j]] = ;
     }
    spfa(s);
    printf("%d", dis[t]);
    return ;
}