Gym 101170I Iron and Coal（BFS + 思维）题解

题意：有一个有向图，有些点是煤，有些点是铁，但不会同时有铁和煤。现在我要从1出发，占领可以到达的点。问最少占领几个点能同时拥有一个煤和一个铁（1不用占领）。

思路：思路很秀啊。我们从1往外bfs，得到所有点到1的最短路dis[i][0]，然后从所有煤跑bfs得到所有点到煤的最短路dis[i][1]，我们再从所有铁跑bfs得到所有点到铁的最短路dis[i][2]，那么dis[i][0] + dis[i][1] + dis[i][2]就是以i为分界点分别前往煤和铁的占领的最小距离。那么答案就是min{ dis[i][0] + dis[i][1] + dis[i][2] }。

代码：

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 10;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, m, k;
int iron[maxn], coal[maxn], vis[maxn];
vector<int> G[maxn], g[maxn];
int dis[maxn][3];
struct node{
    int x;
    int step;
};
void bfs1(){
    for(int i = 1; i <= n; i++) vis[i] = 0;
    queue<node> q;
    while(!q.empty()) q.pop();
    vis[1] = 1;
    dis[1][0] = 0;
    node a, b;
    a.x = 1, a.step = 0;
    q.push(a);
    while(!q.empty()){
        a = q.front();
        q.pop();
        for(int i = 0; i < G[a.x].size(); i++){
            int v = G[a.x][i];
            if(vis[v]) continue;
            vis[v] = 1;
            dis[v][0] = a.step + 1;
            b.x = v, b.step = a.step + 1;
            q.push(b);
        }
    }
}
void bfs2(){
    for(int i = 1; i <= n; i++) vis[i] = 0;
    queue<node> q;
    while(!q.empty()) q.pop();
    node a, b;
    for(int i = 1; i <= n; i++){
        if(iron[i]){
            vis[i] = 1;
            dis[i][1] = 0;
            a.x = i, a.step = 0;
            q.push(a);
        }
    }
    while(!q.empty()){
        a = q.front();
        q.pop();
        for(int i = 0; i < g[a.x].size(); i++){
            int v = g[a.x][i];
            if(vis[v]) continue;
            vis[v] = 1;
            dis[v][1] = a.step + 1;
            b.x = v, b.step = a.step + 1;
            q.push(b);
        }
    }
}
void bfs3(){
    for(int i = 1; i <= n; i++) vis[i] = 0;
    queue<node> q;
    while(!q.empty()) q.pop();
    node a, b;
    for(int i = 1; i <= n; i++){
        if(coal[i]){
            vis[i] = 1;
            dis[i][2] = 0;
            a.x = i, a.step = 0;
            q.push(a);
        }
    }
    while(!q.empty()){
        a = q.front();
        q.pop();
        for(int i = 0; i < g[a.x].size(); i++){
            int v = g[a.x][i];
            if(vis[v]) continue;
            vis[v] = 1;
            dis[v][2] = a.step + 1;
            b.x = v, b.step = a.step + 1;
            q.push(b);
        }
    }
}
int main(){
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; i++){
        iron[i] = coal[i] = 0;
        G[i].clear();
        g[i].clear();
        memset(dis[i], INF, sizeof(dis[i]));
    }
    for(int i = 1; i <= m; i++){
        int u;
        scanf("%d", &u);
        iron[u] = 1;
    }
    for(int i = 1; i <= k; i++){
        int u;
        scanf("%d", &u);
        coal[u] = 1;
    }
    for(int i = 1; i <= n; i++){
        int p, u, v;
        scanf("%d", &p);
        u = i;
        while(p--){
            scanf("%d", &v);
            G[u].push_back(v);
            g[v].push_back(u);
        }
    }
    bfs1();
    bfs2();
    bfs3();
    int ans = INF;
    for(int i = 1; i <= n; i++){
        if(dis[i][0] == INF || dis[i][1] == INF || dis[i][2] == INF) continue;
        ans = min(ans, dis[i][0] + dis[i][1] + dis[i][2]);
    }
    if(ans < INF) printf("%d\n", ans);
    else printf("impossible\n");
    return 0;
}