给出一个图，求两点的曼哈顿距离（|det(x)|+|det(y)|）。题目本身不难，用并查集记录集合及两点间的关系，可我不得不说这题给出数据的方式相当坑爹，硬是把一个在线的问题搞成离线的，就好像活生生把一个直男掰弯一样。
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 #define N 40010 6 #define M 10005 7 int fa[N]; 8 int X[26],Y[26]; 9 struct Dir{ 10 int x,y; 11 }dir[N]; 12 struct Road{ 13 int u,v,x,y; 14 }road[N]; 15 struct Query{ 16 int u,v,id,ans; 17 }qry[M]; 18 int r[M]; 19 bool cmp(int a,int b){ 20 return qry[a].id<qry[b].id; 21 } 22 int find(int s){ 23 if(fa[s] == s) 24 { 25 dir[s].x=dir[s].y=0; 26 return s; 27 } 28 int t = fa[s]; 29 fa[s] = find(t); 30 dir[s].x += dir[t].x; 31 dir[s].y += dir[t].y; 32 return fa[s]; 33 } 34 void comb(int n) 35 { 36 int s,t; 37 Road m; 38 m = road[n]; 39 s = find(m.u); 40 t = find(m.v); 41 if(s == t) return ; 42 fa[t] = s; 43 dir[t].x = dir[m.u].x+m.x-dir[m.v].x; 44 dir[t].y = dir[m.u].y+m.y-dir[m.v].y; 45 } 46 inline int cal(int a,int b){ 47 return abs(dir[a].x-dir[b].x)+abs(dir[a].y-dir[b].y); 48 } 49 int main() 50 { 51 int m,n,k,l,i,j; 52 char ch; 53 Y['N'-'A']=1; 54 Y['S'-'A']=-1; 55 X['W'-'A']=-1; 56 X['E'-'A']=1; 57 scanf("%d%d",&n,&m); 58 for(i = 1; i <= m; i++) 59 { 60 fa[i]=i; 61 scanf("%d%d%d %c",&road[i].u,&road[i].v,&l,&ch); 62 road[i].x = l*X[ch-'A']; 63 road[i].y = l*Y[ch-'A']; 64 } 65 scanf("%d",&k); 66 for(i = 1; i <= k; i++) 67 { 68 r[i] = i; 69 scanf("%d%d%d",&qry[i].u,&qry[i].v,&qry[i].id); 70 } 71 sort(r,r+k,cmp); 72 for(i = j = 1; i <= k; i++) 73 { 74 Query &t = qry[r[i]]; 75 for(;j <= t.id; j++) comb(j); 76 if(find(t.u)!=find(t.v)) 77 t.ans = -1; 78 else t.ans = cal(t.u,t.v); 79 } 80 for(i = 1; i <= k; i++) printf("%d\n",qry[i].ans); 81 return 0; 82 }