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《洛谷P2296 尋找道路》

這題挺好的吧~

先處理出不能走的點,然後再跑最短路的時候去判斷就行。

對於不能走的點:

先建反圖跑最短路,給不能走到終點的點打上標記,然後和這些點相連的點也打上標記。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e4+5;
const int M = 2e5+5;
const LL Mod = 1e9+7;
#define rg register
#define pi acos(-1)
#define
INF 1e9 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; void FRE(){/*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);
*/} int n,m,dis[N],vis[N],s,t; vector<int> e[N],G[N]; void dij() { for(rg int i = 1;i <= n;++i) dis[i] = INF; dis[t] = 0; priority_queue<pii,vector<pii>,greater<pii> >Q; Q.push(pii{0,t}); while(!Q.empty()) { int u = Q.top().second; int d = Q.top().first; Q.pop(); if(d > dis[u]) continue; for(auto v : G[u]) { if(dis[v] > dis[u]+1) { dis[v] = dis[u]+1; Q.push(pii{dis[v],v}); } } } } void dij2() { for(rg int i = 1;i <= n;++i) dis[i] = INF; dis[s] = 0; priority_queue<pii,vector<pii>,greater<pii> >Q; Q.push(pii{0,s}); while(!Q.empty()) { int u = Q.top().second; int d = Q.top().first; Q.pop(); if(d > dis[u]) continue; for(auto v : e[u]) { if(vis[v]) continue; if(dis[v] > dis[u]+1) { dis[v] = dis[u]+1; Q.push(pii{dis[v],v}); } } } } int main() { n = read(),m = read(); for(rg int i = 1;i <= m;++i) { int x,y;x = read(),y = read(); e[x].push_back(y); G[y].push_back(x); } s = read(),t = read(); dij(); for(rg int i = 1;i <= n;++i) { if(dis[i] != INF) continue; vis[i] = 1; for(auto v : G[i]) vis[v] = 1; } if(vis[s]) printf("-1\n"); else { dij2(); if(dis[t] == INF) printf("-1\n"); else printf("%d\n",dis[t]); } system("pause"); }
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