HDU

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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！

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