POJ 1127 Jack Straws

 

Jack Straws

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1824		Accepted: 817

Description

In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned with if various pairs of straws are connected by a path of touching straws. You will be given a list of the endpoints for some straws (as if they were dumped on a large piece of graph paper) and then will be asked if various pairs of straws are connected. Note that touching is connecting, but also two straws can be connected indirectly via other connected straws.

Input

Input consist multiple case,each case consists of multiple lines. The first line will be an integer n (1 < n < 13) giving the number of straws on the table. Each of the next n lines contain 4 positive integers,x1,y1,x2 and y2, giving the coordinates, (x1,y1),(x2,y2) of the endpoints of a single straw. All coordinates will be less than 100. (Note that the straws will be of varying lengths.) The first straw entered will be known as straw #1, the second as straw #2, and so on. The remaining lines of the current case(except for the final line) will each contain two positive integers, a and b, both between 1 and n, inclusive. You are to determine if straw a can be connected to straw b. When a = 0 = b, the current case is terminated.

When n=0,the input is terminated.

There will be no illegal input and there are no zero-length straws.

Output

You should generate a line of output for each line containing a pair a and b, except the final line where a = 0 = b. The line should say simply "CONNECTED", if straw a is connected to straw b, or "NOT CONNECTED", if straw a is not connected to straw b. For our purposes, a straw is considered connected to itself.

Sample Input

7
1 6 3 3 
4 6 4 9 
4 5 6 7 
1 4 3 5 
3 5 5 5 
5 2 6 3 
5 4 7 2 
1 4 
1 6 
3 3 
6 7 
2 3 
1 3 
0 0

2
0 2 0 0
0 0 0 1
1 1
2 2
1 2
0 0

0

Sample Output

CONNECTED 
NOT CONNECTED 
CONNECTED 
CONNECTED 
NOT CONNECTED 
CONNECTED
CONNECTED
CONNECTED
CONNECTED

Source

East Central North America 1994

 

/* http://acm.pku.edu.cn/JudgeOnline/problem?id=1127 计算几何 + 并查集，一开始错以为是动态规划了，忽略了可以向上搜索，汗 */ #include <iostream> #include <memory> #define MAX_N 13 #define MAXV(x, y) ((x) >= (y) ? (x) : (y)) #define MINV(x, y) ((x) <= (y) ? (x) : (y)) using namespace std; int num; struct node { int x1, y1, x2, y2; }data[MAX_N + 1]; int set[MAX_N + 1]; int rank[MAX_N + 1]; //判断第3个点在1,2点构成的线短的哪个方向, -1: 逆时针方向, 1: 顺时针方向 int direct(int x1, int y1, int x2, int y2, int x3, int y3) { return (x3 - x1) * (y2 - y1) - (x2 - x1) * (y3 - y1); } //判断第3个点是否在1,2点构成的线短上 bool onLine(int x1, int y1, int x2, int y2, int x3, int y3) { //cout<<MINV(x1, x2)<<" "<<MAXV(x1, x2)<<" "<<MINV(y1, y2)<<" "<<MAXV(y1, y2); return (MINV(x1, x2) <= x3 && x3 <= MAXV(x1, x2) && MINV(y1, y2) <= y3 && y3 <= MAXV(y1, y2)); } //判断点1,2构成的线短与点3,4构成的线短是否相交 bool intersect(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) { int d1 = direct(x3, y3, x4, y4, x1, y1); int d2 = direct(x3, y3, x4, y4, x2, y2); int d3 = direct(x1, y1, x2, y2, x3, y3); int d4 = direct(x1, y1, x2, y2, x4, y4); if(((d1 > 0 && d2 < 0) || (d1 < 0 && d2 > 0)) && ((d3 > 0 && d4 < 0) || (d3 < 0 && d4 > 0))) return true; if(d1 == 0 && onLine(x3, y3, x4, y4, x1, y1)) return true; else if(d2 == 0 && onLine(x3, y3, x4, y4, x2, y2)) return true; else if(d3 == 0 && onLine(x1, y1, x2, y2, x3, y3)) return true; else if(d4 == 0 && onLine(x1, y1, x2, y2, x4, y4)) return true; else return false; } int find(int pos) { if(pos != set[pos]) set[pos] = find(set[pos]); return set[pos]; } void joinSet(int pos1, int pos2) { if(pos1 == pos2) return; int pre1 = find(pos1); int pre2 = find(pos2); if(pre1 == pre2) return; else { int rank1 = rank[pre1]; int rank2 = rank[pre2]; if(rank1 < rank2) set[pre1] = pre2; else if(rank1 > rank2) set[pre2] = pre1; else { rank[pre1]++; set[pre2] = pre1; } } } void init() { int i; for(i = 1; i <= num; i++) { set[i] = i; rank[i] = 0; } } int main() { int i, n1, n2, pre1, pre2; while(cin>>num && num != 0) { for(i = 1; i <= num; i++) cin>>data[i].x1>>data[i].y1>>data[i].x2>>data[i].y2; init(); //n1 = 4, n2 = 5; //if(intersect(data[n1].x1, data[n1].y1, data[n1].x2, data[n1].y2, data[n2].x1, data[n2].y1, data[n2].x2, data[n2].y2)) // cout<<"hello"<<endl; for(n1 = 1; n1 <= num; n1++) { for(n2 = n1; n2 <= num; n2++) { pre1 = find(n1); pre2 = find(n2); if(pre1 == pre2) continue; else { if(intersect(data[n1].x1, data[n1].y1, data[n1].x2, data[n1].y2, data[n2].x1, data[n2].y1, data[n2].x2, data[n2].y2)) joinSet(n1, n2); } } } while(cin>>n1>>n2 && !(n1 == 0 && n2 == 0)) { pre1 = find(n1); pre2 = find(n2); if(pre1 == pre2) cout<<"CONNECTED"<<endl; else cout<<"NOT CONNECTED"<<endl; } } return 0; }