思路:d[x][y][z]表示以z方向走到(x, y)的转弯次数。
如果用优先队列会超时,因为加入队列的节点太多,无用的节点不能及时出队,会造成MLE,用单调队列即可。
AC代码
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1000 + 5;
map<PI, int>dir;
const int dx[] = {0,0,-1,1};
const int dy[] = {1,-1,0,0};
int n, m, d[maxn][maxn][4], G[maxn][maxn];
struct node{
int x, y, tc, dir;
node() {}
node(int x, int y, int tc, int dir):x(x), y(y), tc(tc), dir(dir) {
}
};
bool bfs(int x1, int y1, int x2, int y2) {
memset(d, inf, sizeof(d));
queue<node>q;
for(int i = 0; i < 4; ++i) {
d[x1][y1][i] = 0;
q.push(node(x1, y1, 0, i));
}
while(!q.empty()) {
node p = q.front(); q.pop();
int x = p.x, y = p.y, tc = p.tc, dir = p.dir;
if(tc > d[x][y][dir]) continue;
for(int i = 0; i < 4; ++i) {
int px = x + dx[i], py = y + dy[i];
if(px < 0 || py < 0 || px >= n || py >= m) continue;
int pt = tc;
if(i != dir) ++pt; //发生偏转
if(pt > 2) continue;
if(px == x2 && py == y2) return true;
if(G[px][py]) continue;
if(pt < d[px][py][i]) {
d[px][py][i] = pt;
q.push(node(px, py, pt, i));
}
}
}
return false;
}
int main() {
int q;
while(scanf("%d%d", &n, &m) == 2 && n && m) {
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
scanf("%d", &G[i][j]);
scanf("%d", &q);
int x1, y1, x2, y2;
while(q--) {
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
--x1, --y1, --x2, --y2;
if( G[x1][y1] && G[x2][y2] && G[x1][y1] == G[x2][y2] && bfs(x1, y1, x2, y2)) printf("YES\n");
else printf("NO\n");
}
}
return 0;
}
如有不当之处欢迎指出!