标签：printf color front iss mission std code return gre
魔王每t分钟回地牢视察一次。若发现Ignatius不在原位置便把他拎回去。经过若干次的尝试，Ignatius已画出整个地牢的地图。如今请你帮他计算是否能再次成功逃亡。仅仅要在魔王下次视察之前走到出口就算离开地牢，假设魔王回来的时候刚好走到出口或还未到出口都算逃亡失败。
4 5 17 @A.B. a*.*. *..*^ c..b* 4 5 16 @A.B. a*.*. *..*^ c..b*
16 -1
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 25
using namespace std;
int vis[maxn][maxn][1 << 11];
char map[maxn][maxn];
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
struct node{
int x, y ,step, key;
};
int n, m, t, sx, sy;
bool check(node a){
if(a.x >= 0 && a.x < n && a.y >= 0 && a.y < m && map[a.x][a.y] != ‘*‘)
return true;
else
return false;
}
int bfs(){
node now, next;
queue<node >q;
now.x = sx;
now.y = sy;
now.step = 0;
now.key = 0;
vis[now.x][now.y][now.key] = true;
q.push(now);
while(!q.empty()){
now = q.front();
q.pop();
if(map[now.x][now.y] == ‘^‘ && now.step < t){
return now.step;
}
if(now.step > t) continue;
for(int i = 0; i < 4; ++i){
next.x = now.x + dir[i][0];
next.y = now.y + dir[i][1];
next.step = now.step + 1;
if(check(next)){
if(map[next.x][next.y] >= ‘a‘ && map[next.x][next.y] <= ‘z‘){//钥匙
next.key = now.key | (1 << (map[next.x][next.y] - ‘a‘));//获得这个钥匙
if(!vis[next.x][next.y][next.key]){
vis[next.x][next.y][next.key] = 1;
q.push(next);
}
}
else if(map[next.x][next.y] >= ‘A‘ && map[next.x][next.y] <= ‘Z‘){//门
next.key = now.key;
if(next.key & (1 << (map[next.x][next.y] - ‘A‘))){//拥有这个门的钥匙
if(!vis[next.x][next.y][next.key]){
vis[next.x][next.y][next.key] = 1;
q.push(next);
}
}
}
else {//路
next.key = now.key;
if(!vis[next.x][next.y][next.key]){
vis[next.x][next.y][next.key] = 1;
q.push(next);
}
}
}
}
}
return -1;
}
int main (){
while(scanf("%d%d%d", &n, &m, &t) != EOF){
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; ++i){
scanf("%s", map[i]);
for(int j = 0; j < m; ++j)
if(map[i][j] == ‘@‘)
sx = i, sy = j;
}
int ans;
ans = bfs();
printf("%d\n", ans);
}
return 0;
}
HDU 1429--胜利大逃亡(续)【BFS && 状态压缩】
标签：printf color front iss mission std code return gre
原文地址：http://www.cnblogs.com/gccbuaa/p/6885869.html
