UVA - 321 The New Villa

题目大意：别墅有 r 个房间，d 扇门， s 个开关，判断是否能从编号 1 的房间走到编号 r 的房间，若能打印步骤。两个房间之间有门的才可移动，在移动时注意不能进入没有亮灯的房间。
解题思路：二进制表示房间状态，bfs 查找递归输出。

#include<iostream> #include<cstdio>#include<cmath>#include<string.h>#include<stdlib.h>#include<algorithm>#include<queue>#include<map>using namespace std;int r, d, s;int cnt = 0;int vis[15][2000];int pre[15][2000];struct node {    int step, now, state;};queue<node> q;int door[15][15], light[15][15];node now, nt;void bfs() {    int n, sta, t;    while (!q.empty() && !vis[r][1]) {        now = q.front(); q.pop();        n = now.now;        sta = now.state * 100 + n;        nt.step = now.step + 1;        for (int i = 1; i <= r; i++) {            if (!light[n][i] || i == n) continue;            t = now.state ^ (1 << (r-i));            if (vis[n][t]) continue;            vis[n][t] = nt.step;            pre[n][t] = sta;            nt.state = t;            nt.now = n;            q.push(nt);        }        for (int i = 1; i <= r; i++) {            if (!door[n][i]) continue;            if (!((1<<(r-i)) & now.state) || vis[i][now.state]) continue;            vis[i][now.state] = nt.step;            pre[i][now.state] = sta;            nt.state = now.state;            nt.now = i;            q.push(nt);        }    }}void output(int n, int sta) {    int ln, ls, t = pre[n][sta];    if (!t) return;    ln = t % 100;    ls = t / 100;    output(ln, ls);    if(ln != n)printf("- Move to room %d.\n", n);      else {        int t=ls^sta, i, flag;        for(i = 0; i < r&&!(t&1); i++, t>>=1);          flag = ls&(1<<i);          printf("- Switch %s light in room %d.\n",flag?"off":"on",r-i);    }  }int main() {    while (scanf("%d%d%d", &r, &d, &s) != EOF && r+d+s) {        memset(door, 0, sizeof(door));        memset(light, 0, sizeof(light));        memset(vis, 0, sizeof(vis));        int a, b;        for (int i = 0; i < d; i++) {            scanf("%d%d", &a, &b);            door[a][b] = door[b][a] = 1;        }        for (int i = 0; i < s; i++) {            scanf("%d%d", &a, &b);            light[a][b] = 1;        }        while (!q.empty()) q.pop();        int i = 1 << (r - 1);        vis[1][i] = 1;        pre[1][i] = 0;        now.state = i;        now.now = 1;        now.step = 1;        q.push(now);        printf("Villa #%d\n", ++cnt);        bfs();        if(vis[r][1]) {            printf("The problem can be solved in %d steps:\n",vis[r][1]-1);            output(r,1);        }        else printf("The problem cannot be solved.\n");          printf("\n");    }return 0;}