ZOJ3761（并查集+树的遍历)
Edward think a game of billiards is too long and boring. So he invented a new game called Easy billiards.
Easy billiards has N balls on a brimless rectangular table in the beginning, and your goal is try to make the number of balls on the table as least as possible by several hit under the following rules:
1: The direction you hit the balls should parallel to the tables border.
2: If ball A crashed into ball B, ball B will moves in the same direction of ball A before the crashing, and ball A will stop in the place of ball B before the crashing.
3: If ball C is moving and there are no balls in front of ball C, it will runs out of the tables border, that means ball C is out of the table.
4: You can choose arbitrary ball on the table to hit, but on a hit, you can't let the ball you choose to hit runs out of the tables border. In another word, a ball could runs out of the table if and only if it was crashed by another ball in a hitting.
Now, Edward wants to know the least number of balls remained on the table after several hits, and how.
Input
There are multiple test cases. For each test cases, in the first line, there is an integer N, which means the number of the balls on the table. There are following N lines, each line contains two integers Xi and Yi, which means the coordinate of ball I. (0<=N<=2000, 0<=Xi, Yi<=10^8)
Output
For each test cases, you should output the least number of balls on the first line. And you should output several lines to show the order of hits following the first line, each line should contains the coordinate of the ball you choose to hit and the direction you hit. (LEFT,RIGHT,UP,DOWN).
Sample Input
4 0 0 2 0 4 0 2 2 9 1 1 2 1 3 1 1 2 2 2 3 2 1 3 2 3 3 3
Sample output
1
(2, 2) DOWN
(4, 0) LEFT
(2, 0) LEFT
1
(1, 3) DOWN
(1, 2) DOWN
(2, 3) DOWN
(2, 2) DOWN
(3, 3) DOWN
(3, 2) DOWN
(3, 1) LEFT
(2, 1) LEFT
题意：一块区域内，通过撞球来使区域内球数最少。撞球规则：A球在B球右边而B球右边没球存在，这时，向右撞击A球，A球会撞击B球,A球会停在B球的位置上，B球则被撞出这片区域。
思路：先并查集得到最少撞球数。再DFS遍历树得到撞球方向。
收获：注意标记位置的放法，用链式前向星建无向图。
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cstdlib> 5 using namespace std; 6 #define maxn 10000 7 #define maxm 1000000 8 int n,num; 9 struct Node 10 { 11 int x; 12 int y; 13 }; 14 Node node[maxn]; 15 int root[maxn]; 16 void init_root() 17 { 18 for(int i=0;i<n;i++) 19 root[i]=i; 20 } 21 int findroot(int x) 22 { 23 if(x!=root[x]) 24 { 25 root[x]=findroot(root[x]); 26 } 27 return root[x]; 28 } 29 void merge_(int a,int b) 30 { 31 int x=findroot(a); 32 int y=findroot(b); 33 if(x==y) 34 return; 35 else 36 root[y]=x; 37 } 38 struct Edge 39 { 40 int u,v,next; 41 }; 42 Edge edge[maxm]; 43 int head[maxm]; 44 int vis[maxm]; 45 void init_edge() 46 { 47 num=0; 48 memset(vis,0,sizeof(vis)); 49 memset(head,-1,sizeof(head)); 50 } 51 void addedge(int u,int v) 52 { 53 edge[num].u=u; 54 edge[num].v=v; 55 edge[num].next=head[u]; 56 head[u]=num++; 57 edge[num].u=v; 58 edge[num].v=u; 59 edge[num].next=head[v]; 60 head[v]=num++; 61 } 62 void dfs(int pre,int u) 63 { 64 //printf("%d%d\n",node[u].x,node[u].y); 65 vis[u]=1; 66 for(int i=head[u];i!=-1;i=edge[i].next) 67 { 68 int v=edge[i].v; 69 if(!vis[v]) 70 { 71 dfs(u,v); 72 //vis[v]=1; 73 } 74 } 75 if(pre!=-1) 76 { 77 printf("(%d, %d) ",node[u].x,node[u].y); 78 if(node[pre].x>node[u].x) 79 printf("RIGHT\n"); 80 if(node[pre].x<node[u].x) 81 printf("LEFT\n"); 82 if(node[pre].x==node[u].x) 83 { 84 if(node[pre].y>node[u].y) 85 printf("UP\n"); 86 else 87 printf("DOWN\n"); 88 } 89 } 90 91 } 92 int main() 93 { 94 while(~scanf("%d",&n)) 95 { 96 for(int i=0;i<n;i++) 97 scanf("%d%d",&node[i].x,&node[i].y); 98 init_root(); 99 init_edge(); 100 for(int i=0;i<n;i++) 101 for(int j=0;j<i;j++) 102 { 103 if(node[i].x==node[j].x||node[i].y==node[j].y) 104 { 105 merge_(i,j); 106 addedge(i,j); 107 } 108 } 109 memset(vis,0,sizeof(vis)); 110 int ans=0; 111 for (int i=0;i<n;i++) 112 if (root[i]==i) ans++; 113 printf("%d\n",ans); 114 for(int i=0;i<n;i++) 115 if(root[i]==i) 116 { 117 dfs(-1,i); 118 } 119 } 120 return 0; 121 }
