d[i][j][k]代表在(i,j)这个坐标点，钥匙集合为二进制k的最短路。
跑一遍Dijkstra即可。
#include<bits/stdc++.h> using namespace std; const int N=15; const int dx[]={-1,0,1,0}; const int dy[]={0,1,0,-1}; const int INF=0x3f3f3f3f; int n,m,k,mp[110][110]; vector<int> ys[110]; struct node{ int x,y,ys; int d; node() {} node(int x,int y,int ys,int d) : x(x),y(y),ys(ys),d(d) {} bool operator < (const node &rhs) const { return d>rhs.d; } }; int id(int x,int y) { return (x-1)*m+y; } int d[N][N][1<<12]; bool vis[N][N][1<<12]; priority_queue<node> q; int Dijkstra() { int ret=INF; memset(d,0x3f,sizeof(d)); memset(vis,0,sizeof(vis)); int tmp=0; for (int i=0;i<ys[1].size();i++) tmp|=ys[1][i]; d[1][1][tmp]=0; q.push(node(1,1,tmp,0)); while (!q.empty()) { node u=q.top(); q.pop(); if (u.x==n && u.y==m) ret=min(ret,u.d); if (vis[u.x][u.y][u.ys]) continue; vis[u.x][u.y][u.ys]=1; for (int i=0;i<4;i++) { int nx=u.x+dx[i],ny=u.y+dy[i]; if (nx<1 || nx>n || ny<1 || ny>m || mp[id(u.x,u.y)][id(nx,ny)]==0) continue; if (mp[id(u.x,u.y)][id(nx,ny)]>0 && !(u.ys&(1<<mp[id(u.x,u.y)][id(nx,ny)]))) continue; int tmp=u.ys; for (int j=0;j<ys[id(nx,ny)].size();j++) tmp|=(1<<ys[id(nx,ny)][j]); if (d[u.x][u.y][u.ys]+1<d[nx][ny][tmp]) { d[nx][ny][tmp]=d[u.x][u.y][u.ys]+1; q.push(node(nx,ny,tmp,d[nx][ny][tmp])); } } } return ret==INF ? -1 : ret; } int main() { scanf("%d%d%d",&n,&m,&k); int t; scanf("%d",&t); memset(mp,-1,sizeof(mp)); for (int i=1;i<=t;i++) { int x1,y1,x2,y2,p; scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&p); mp[id(x1,y1)][id(x2,y2)]=mp[id(x2,y2)][id(x1,y1)]=p; } scanf("%d",&t); for (int i=1;i<=t;i++) { int x,y,p; scanf("%d%d%d",&x,&y,&p); ys[id(x,y)].push_back(p); } cout<<Dijkstra()<<endl; return 0; }