搜索小练#9

题目

今天没写完

晚上倒是有一段时间

不过我去弄Atom了......

明天计划:

切掉这个题 把以前做过的搜索题复习一下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N = 	66, INF = 214748364;

int T, n, m, res = INF;
int a[N][N], flag;
char s[N];

int dx[3] = {0, 0, 1, -1};
int dy[3] = {1, -1, 0, 0};
int vs[N][N], vt[N][N];
int sdis[N][N], tdis[N][N];

struct node {int x, y;};

inline int bfs (int curx, int cury, int curs, int curt) {
	// if (x == nx && y == ny) calc();
	queue<node>qs; queue<node>qt;
  node l, r;
	l.x = curx, l.y = cury, r.x = curs, r.y = curt;
	qs.push(l); qt.push(r);
	memset (vs, 0, sizeof vs), memset (vt, 0, sizeof vt);
	memset (sdis, 0, sizeof sdis), memset (tdis, 0, sizeof tdis);
	int ans(0);
	while (qs.size()) {
		node tmp = qs.front(); qs.pop();
		for (int i = 0; i < 3; ++ i) {
			node nows;
			nows.x = tmp.x+dx[i], nows.y = tmp.y+dy[i];
			if ((nows.x > n)||(nows.x < 1)||(nows.y > m)||(nows.y < 1)) continue;
			if (!vs[nows.x][nows.y] && a[nows.x][nows.y]){
				vs[nows.x][nows.y] = 1;
				qs.push(nows);
				sdis[nows.x][nows.y] = sdis[tmp.x][tmp.y]+1;

			}
		}
	}
	while (qt.size()) {
		node tmp = qt.front(); qt.pop();
		for (int i = 0; i < 3; ++ i) {
			node nows;
			nows.x = tmp.x+dx[i], nows.y = tmp.y+dy[i];
			if ((nows.x > n)||(nows.x < 1)||(nows.y > m)||(nows.y < 1)) continue;
			if (!vt[nows.x][nows.y] && a[nows.x][nows.y]){
				vt[nows.x][nows.y] = 1;
				qt.push(nows);
				tdis[nows.x][nows.y] = tdis[tmp.x][tmp.y]+1;
			}
		}
	}
	for (int i = 1; i <= n; ++ i)
		for (int j = 1; j <= m; ++ j) {
			if ((!vs[i][j] || !vt[i][j]) && a[i][j]) {
				flag = true;
				return;
			}
		}
}

int main () {
	scanf ("%d", &T);
	while (T --> 0) {
		flag = false, res = 0;
		scanf ("%d%d", &n, &m);
		for (int i = 1; i <= n; ++ i) {
			scanf ("%s", s+1);
			for (int j = 1; j <= m; ++ j)
				a[i][j] = (s[j] == '#') ? 1 : 0;
		}
		for (int i = 1; i <= n; ++ i)
			for (int j = 1; j <= m; ++ j)
				if (a[i][j])
		for (int ni = 1; ni <= n; ++ ni)
			for (int nj = 1; nj <= m; ++ nj)
				if (a[ni][nj])
					bfs (i, j, ni, nj);
		if (flag) cout << "-1" << '\n';
		else cout << res << '\n';
	}
	return 0;
}