题意: 给定起点终点, 找一条从起点到终点的最短路径使路上的每个点都能有路径到达终点.
我们先反着建一遍图, 然后从终点开始 bfs 一遍图, 标记所有终点可以到达的点. 然后再枚举一遍点, 如果这个点是终点没法到达的点, 那么再枚举这个点所能连接的所有点, 如果这些点是终点可以到达的, 那么这些点就是不符合条件的. 最后在符合条件的点上做一遍最短路.
- #include <queue>
- #include <cstdio>
- #include <cstring>
- #include <iostream>
- #include <algorithm>
- using namespace std;
- const int maxn = 200010;
- int start, end, n, m, u[maxn], v[maxn], dis[maxn];
- bool vis1[maxn], vis2[maxn], vis[maxn];
- struct edge{
- int from, to, next, len;
- }e1[maxn<<2], e[maxn<<2];
- int cnt1, cnt, head1[maxn], head[maxn];
- void add1(int u, int v, int w)
- {
- e1[++cnt1].from = u;
- e1[cnt1].len = w;
- e1[cnt1].to = v;
- e1[cnt1].next = head1[u];
- head1[u] = cnt1;
- }
- void add(int u, int v, int w)
- {
- e[++cnt].from = u;
- e[cnt].len = w;
- e[cnt].to = v;
- e[cnt].next = head[u];
- head[u] = cnt;
- }
- queue<int> q1, q;
- void bfs()
- {
- q1.push(end), vis1[end] = 1;
- while(!q1.empty())
- {
- int now = q1.front();
- q1.pop();
- for(int i = head1[now]; i != -1; i = e1[i].next)
- {
- if(!vis1[e1[i].to])
- {
- vis1[e1[i].to] = 1;
- q1.push(e1[i].to);
- }
- }
- }
- memcpy(vis2, vis1, sizeof(vis1));
- for(int i = 1; i <= n; i++)
- {
- if(vis1[i] == 0)
- for(int j = head1[i]; j != -1; j = e1[j].next)
- {
- if(vis2[e1[j].to] == 1)
- vis2[e1[j].to] = 0;
- }
- }
- }
- void SPFA()
- {
- q.push(start);
- vis[start] = 1, dis[start] = 0;
- while(!q.empty())
- {
- int now = q.front();
- q.pop();
- for(int i = head[now]; i != -1; i = e[i].next)
- {
- if(dis[e[i].to]> dis[now] + e[i].len)
- {
- dis[e[i].to] = dis[now] + e[i].len;
- if(!vis[e[i].to])
- {
- q.push(e[i].to);
- }
- }
- }
- }
- }
- int main()
- {
- memset(head1, -1, sizeof(head1));
- memset(head, -1, sizeof(head));
- scanf("%d%d",&n,&m);
- for(int i = 1; i <= n; i++) dis[i] = 233333333;
- for(int i = 1; i <= m; i++)
- {
- scanf("%d%d",&u[i],&v[i]);
- if(u[i] != v[i])
- add1(v[i], u[i], 1);
- }
- scanf("%d%d",&start,&end);
- bfs();
- for(int i = 1; i <= m; i++)
- {
- if(u[i] != v[i] && vis2[u[i]] != 0 && vis2[v[i]] != 0)
- add(u[i], v[i], 1);
- }
- SPFA();
- if(dis[end] == 233333333)
- {
- printf("-1\n");
- return 0;
- }
- else
- printf("%d\n",dis[end]);
- return 0;
- }
来源: http://www.bubuko.com/infodetail-2766648.html