HDU 4771

http://acm.hdu.edu.cn/showproblem.php?pid=4771

给一个地图，@是起点，给一些物品坐标，问取完所有物品的最小步数，不能取完输出-1

物品数最多只有四个，状态压缩一下bfs即可

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>

using namespace std;

int n, m, k;

char G[105][105];
int vis[105][105][1<<4];

struct p {
int x, y, key, step;
};

int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};

p now;

int bfs() {
queue <p> q;
now.step = 0;
q.push(now);
vis[now.x][now.y][0] = 1;
while(!q.empty()) {
p u = q.front();
q.pop();
for(int i = 0; i < 4; i++) {
int xx = u.x + dx[i];
int yy = u.y + dy[i];
if(xx < 0 || xx >= n || yy < 0 || yy >= m) continue;
if(G[xx][yy] == '#') continue;
p next;
if(G[xx][yy] >= '0' && G[xx][yy] <= '3') {
if(!vis[xx][yy][u.key]) {
vis[xx][yy][u.key|(1<<(G[xx][yy]-'0'))] = 1;
next.x = xx, next.y = yy, next.key = u.key|(1<<(G[xx][yy]-'0')), next.step = u.step + 1;
if(next.key == (1<<k)-1) return next.step;
q.push(next);
}
}
else {
if(!vis[xx][yy][u.key]) {
vis[xx][yy][u.key] = 1;
next.x = xx, next.y = yy, next.key = u.key, next.step = u.step + 1;
q.push(next);
}
}
}
}
return -1;
}

int main() {
while(~scanf("%d%d", &n, &m)) {
if(!n && !m) break;
for(int i = 0; i < n; i++)
scanf("%s", G[i]);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(G[i][j] == '@')
now.x = i, now.y = j;
scanf("%d", &k);
int flag = 1;
now.key = 0;
memset(vis, 0, sizeof(vis));
for(int i = 0; i < k; i++) {
int x, y;
scanf("%d%d", &x, &y);
if(G[x-1][y-1] == '#') flag = 0;
else if(G[x-1][y-1] == '@') {
G[x-1][y-1] = i + '0';
vis[x-1][y-1][1<<i] = 1;
}
else G[x-1][y-1] = i + '0';
}
if(!flag) {
puts("-1");
continue;
}
printf("%d\n", bfs());
}
return 0;
}

View Code