「网络流 24 题」孤岛营救问题

题目链接

题解

• 由于$p$很小，所以可以考虑将当前有哪些钥匙用二进制压缩，作为一层状态，这样最多有$n*m*2^p$个状态
• 这样处理过后在图上$bfs$求解即可，注意一个房间可以能有多个钥匙

查看代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> p;
const int maxn = 1e5+5;
const int mod = 1e9+7;
ll qpow(ll a,ll b){ll res=1;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int d[11][11][2100];
int key[11][11];
bool vis[11][11][2100];
map<pair<p,p>,int>mp;
int n,m,k;
struct node
{
    int x,y,sta,cost;
    node(int a,int b,int c,int d):x(a),y(b),sta(c),cost(d){}
};
int dir[][2]={0,1,0,-1,1,0,-1,0};
void bfs()
{
    queue<node>q;
    q.push(node(1,1,0,0));
    vis[1][1][0]=1;
    while(!q.empty()){
        node now = q.front();
        q.pop();
        if(now.x==n&&now.y==m){
            printf("%d\n",now.cost);
            return;
        }
        for(int i = 0;i < 4;++i){
            int x = now.x+dir[i][0];
            int y = now.y+dir[i][1];
            if(x<1||x>n||y<1||y>m)continue;
            int tmp=-1;
            auto ite = mp.find(make_pair(make_pair(now.x,now.y),make_pair(x,y)));
            if(ite!=mp.end()){
                tmp = ite->second;
                if(!((now.sta>>tmp)&1))
                continue;
            }
            int s = now.sta;
            if(key[x][y]!=0)
            s = now.sta|(key[x][y]);
            if(vis[x][y][s])continue;
            vis[x][y][s]=1;
            q.push(node(x,y,s,now.cost+1));
        }
    }
    printf("-1\n");
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("simple.in", "r", stdin);
    freopen("simple.out", "w", stdout);
#endif
    scanf("%d%d%d",&n,&m,&k);
    int p;
    scanf("%d",&p);
    for(int i = 1,a,b,c,d,e;i <= p;++i){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);
        mp[make_pair(make_pair(a,b),make_pair(c,d))]=e;
        mp[make_pair(make_pair(c,d),make_pair(a,b))]=e;
    }
    int fd;
    scanf("%d",&fd);
    for(int i = 1,a,b,c;i <= fd;++i){
        scanf("%d%d%d",&a,&b,&c);
        key[a][b]|=(1<<c);
    }
    bfs();
    return 0;
}