思路：
将边当作点，这样建墙的时候点的坐标变换成x' = 2*x-1, y' = 2*y-1,t' = 2*t, 建门同理，但是因为门只要开一格就行所以t还是1。nemo的位置变换后是下取整x,y坐标各＋1，
然后 1代表墙，2代表门，4代表nemo。这题坑在nemo的范围不知道，所以不能直接在开的建图的数组里找，我就这样RE了好多次，不在图的范围里的直接可以判断输出0。然后用优先队列存经过的点，按经过的门次数升序排序。
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #include <vector> 5 #include <cmath> 6 #include <cstring> 7 #include <string> 8 #include <algorithm> 9 using namespace std; 10 11 int arr[500][500], minStep; 12 bool vis[500][500]; 13 int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 14 struct node 15 { 16 int x, y,step; 17 node(){ 18 x = y = step = 0; 19 } 20 bool operator == (node that){ 21 if(that.x == x && that.y == y) return true; 22 return false; 23 } 24 bool operator < (const node &that) const 25 { 26 return step > that.step; 27 } 28 }; 29 30 void init() 31 { 32 memset(arr, 0, sizeof(arr)); 33 memset(vis, 0, sizeof(vis)); 34 minStep = 160000; 35 } 36 void build_wall(int xx, int yy, int dd, int tt) 37 { 38 int x = 2*xx-1, y = 2*yy-1; 39 if(dd) 40 for(int i = y; i <= y+2*tt; i++) 41 arr[x][i] = 1; 42 else 43 for(int i = x; i <= x+2*tt; i++) 44 arr[i][y] = 1; 45 } 46 void door(int xx, int yy, int dd) 47 { 48 int x = 2 * xx -1, y = 2 * yy -1; 49 if(dd) arr[x][y+1] = 2; 50 else arr[x+1][y] = 2; 51 } 52 bool checkBound(int x, int y) 53 { 54 if(x < 0||y < 0||x>400||y>400) return false; 55 return true; 56 } 57 void findNemo() 58 { 59 node nd, nd2; 60 vis[0][0] = 1; 61 priority_queue<node> q; 62 q.push(nd); 63 while(!q.empty()){ 64 nd = q.top(); q.pop(); 65 for(int i = 0; i < 4; i++) { 66 int x = dir[i][0] + nd.x, y = dir[i][1] + nd.y; 67 if(checkBound(x, y) && arr[x][y] != 1 && !vis[x][y]) { 68 nd2.step = nd.step; nd2.x = x; nd2.y = y; 69 vis[x][y] = 1; 70 if(arr[x][y] == 2) nd2.step++; 71 q.push(nd2); 72 if(arr[x][y] == 4) { 73 vis[x][y] = 0; 74 minStep = nd2.step; 75 return; 76 } 77 } 78 } 79 } 80 } 81 void solve() 82 { 83 int n, m; 84 while(scanf("%d%d", &m, &n) == 2 && n != -1 && m != -1) { 85 init(); 86 int x, y, d, t; 87 for(int i = 0; i < m; i++){ 88 scanf("%d%d%d%d", &x, &y, &d, &t); 89 build_wall(x, y, d, t); 90 } 91 92 for(int i = 0; i < n; i++){ 93 scanf("%d%d%d", &x, &y, &d); 94 door(x, y, d); 95 } 96 double f1, f2; 97 scanf("%lf%lf", &f1, &f2); 98 node nd; nd.x = 2*(int)floor(f1); nd.y = 2*(int)floor(f2); 99 if(checkBound(nd.x, nd.y)){ 100 arr[nd.x][nd.y] = 4; 101 findNemo(); 102 if(minStep == 160000) printf("-1 "); 103 else printf("%d ", minStep); 104 } 105 else printf("0 "); 106 } 107 } 108 int main() 109 { 110 solve(); 111 return 0; 112 }