分析
没事干不要乱建图
隔壁白银莲花池是因为不能有零环才重新建图的
这道题，直接01走起就好
听说是什么01BFS
0插队首，1插队尾，时间复杂度，可以自己算，也可以造个大数据
莫名RE，一直以为应该是TLE来着
代码
1 /************************** 2 User:Mandy.H.Y 3 Language:c++ 4 Problem:luogu4554 5 Algorithm: 6 Date:2019.9.4 7 **************************/ 8 9 #include<bits/stdc++.h> 10 11 using namespace std; 12 13 const int maxn = 505; 14 const int maxm = 1e6 + 5; 15 16 int n,m; 17 int dis[maxn][maxn]; 18 char a[maxn][maxn]; 19 bool vis[maxn][maxn]; 20 21 struct Node{ 22 int x,y; 23 Node(int _x = 0,int _y = 0):x(_x),y(_y){} 24 }q[300005],s,t; 25 int l,r; 26 27 int dx[5] = {0,0,1,-1}; 28 int dy[5] = {1,-1,0,0}; 29 30 template<class T>inline void read(T &x){ 31 x = 0;bool flag = 0;char ch = getchar(); 32 while(!isdigit(ch)) flag |= ch == '-',ch = getchar(); 33 while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar(); 34 if(flag) x = -x; 35 } 36 37 template<class T>void putch(const T x){ 38 if(x > 9) putch(x / 10); 39 putchar(x % 10 | 48); 40 } 41 42 template<class T>void put(const T x){ 43 if(x < 0) putchar('-'),putch(-x); 44 else putch(x); 45 } 46 47 void file(){ 48 freopen("4554.in","r",stdin); 49 // freopen("4554.out","w",stdout); 50 } 51 52 void readdata(){ 53 for(int i = 0;i < n; ++ i) scanf("%s",a[i]); 54 read(s.x);read(s.y); 55 read(t.x);read(t.y); 56 } 57 58 void SPFA(){ 59 memset(vis,0,sizeof(vis)); 60 memset(dis,0x3f3f3f3f,sizeof(dis)); 61 l = r = 0; 62 q[r++] = s; 63 dis[s.x][s.y] = 0; 64 while(l != r){ 65 Node u = q[l]; 66 int x = u.x; 67 int y = u.y; 68 ++l; 69 l = l % 260000; 70 int rr = (r - 1 + 260000) % 260000; 71 if(l != r && dis[q[l].x][q[l].y] > dis[q[rr].x][q[rr].y]) swap(q[l],q[rr]); 72 vis[x][y] = 0; 73 for(int i = 0;i < 4; ++ i){ 74 int nx = x + dx[i]; 75 int ny = y + dy[i]; 76 if(nx >= n || ny >= m || nx < 0 || ny < 0) continue; 77 int w; 78 if(a[x][y] == a[nx][ny]) w = 0; 79 else w = 1; 80 if(dis[nx][ny] > dis[x][y] + w){ 81 dis[nx][ny] = dis[x][y] + w; 82 if(!vis[nx][ny]){ 83 q[r++] = Node(nx,ny); 84 vis[nx][ny] = 1; 85 r %= 260000; 86 int rr = (r - 1 + 260000) % 260000; 87 if(l != r && dis[q[l].x][q[l].y] > dis[q[rr].x][q[rr].y]) swap(q[l],q[rr]); 88 } 89 90 } 91 } 92 } 93 } 94 95 void work(){ 96 97 SPFA(); 98 99 put(dis[t.x][t.y]); 100 putchar(' '); 101 } 102 103 int main(){ 104 // file(); 105 while(1){ 106 scanf("%d%d",&n,&m); 107 if((!n) && (!m)) break; 108 readdata(); 109 work(); 110 } 111 return 0; 112 }