http://acm.hdu.edu.cn/showproblem.php?pid=5094
bfs,vis[x][y][z],z表示钥匙的状态,用二进制来表示,key[x][y]储存当前位置钥匙的二进制表示。
注意起始点有钥匙的情况。
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<algorithm> #include<queue> using namespace std; struct point { int x,y,counts,mykey; }start; int n,m,p,k,s,flag,door[55][55][55][55],key[55][55],vis[55][55][2048],dir[4][2] = {-1,0,0,-1,1,0,0,1}; int main() { while(~scanf("%d%d%d",&n,&m,&p)) { memset(door,-1,sizeof(door)); memset(key,0,sizeof(key)); memset(vis,0,sizeof(vis)); scanf("%d",&k); int x1,y1,x2,y2,type; while(k--) { scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&type); door[x1][y1][x2][y2] = type; door[x2][y2][x1][y1] = type; } scanf("%d",&s); int x,y,keytype; while(s--) { scanf("%d%d%d",&x,&y,&keytype); key[x][y] |= 1<<(keytype-1); } queue<point> q; start.x = 1; start.y = 1; start.counts = 0; start.mykey = key[1][1]; vis[1][1][start.mykey] = 1; q.push(start); flag = 1; while(!q.empty()) { point now = q.front(); q.pop(); if(now.x == n && now.y == m) { flag = 0; printf("%d\n",now.counts); break; } for(int i = 0;i < 4;i++) { point temp; temp.x = now.x+dir[i][0]; temp.y = now.y+dir[i][1]; if(temp.x < 1 || temp.x > n || temp.y < 1 || temp.y > m) continue; if(door[now.x][now.y][temp.x][temp.y] != -1 &&(now.mykey & 1<<(door[now.x][now.y][temp.x][temp.y]-1)) == 0) continue; temp.counts = now.counts+1; temp.mykey = now.mykey | key[temp.x][temp.y]; if(vis[temp.x][temp.y][temp.mykey]) continue; vis[temp.x][temp.y][temp.mykey] = 1; q.push(temp); } } if(flag) printf("-1\n"); } }
时间: 2024-10-13 11:35:50