HDOJ1558 线段相交问题+并查集处理
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Segment set
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5104 Accepted Submission(s): 1967
Problem Description
A segment and all segments which are connected with it compose a segment set. The size of a segment set is the number of segments in it. The problem is to find the size of some segment set.
Input
In the first line there is an integer t - the number of test case. For each test case in first line there is an integer n (n<=1000) - the number of commands.
There are two different commands described in different format shown below:
P x1 y1 x2 y2 - paint a segment whose coordinates of the two endpoints are (x1,y1),(x2,y2).
Q k - query the size of the segment set which contains the k-th segment.
k is between 1 and the number of segments in the moment. There is no segment in the plane at first, so the first command is always a P-command.
Output
For each Q-command, output the answer. There is a blank line between test cases.
Sample Input
1
10
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
P 2.00 3.00 3.00 1.00
Q 1
Q 3
P 1.00 4.00 8.00 2.00
Q 2
P 3.00 3.00 6.00 -2.00
Q 5
Sample Output
1
2
2
2
5
Author
LL
Source
HDU 2006-12 Programming Contest
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才刚做计算几何的题没多久就遇到了并查集的内容,看来并查集真要好好掌握啊!!
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;const int maxn = 1005;struct point{ double x,y;};struct segment{ point s,e;}a[maxn];int i,j,k,n,tot;int f[maxn],ans[maxn];char cc;double product(point &a, point &b, point &c){ double x1,y1,x2,y2; x1 = a.x - c.x; y1 = a.y - c.y; x2 = b.x - c.x; y2 = b.y - c.y; return (x1*y2 - x2*y1);}bool intersect(segment &a, segment &b){ if (/*max(a.s.x,a.e.x)>=min(b.s.x,b.e.x) && max(a.s.y,a.e.y)>=min(b.s.y,b.e.y) && max(b.s.x,b.e.x)>=min(a.s.x,a.e.x) && max(b.s.y,b.e.y)>=min(a.s.y,a.e.y) &&8*/ product(a.s,a.e,b.s)*product(a.s,a.e,b.e)<=0 && product(b.s,b.e,a.s)*product(b.s,b.e,a.e)<=0 ) return 1; return 0;}int Find(int x) { if (f[x]==x) return x; return f[x] = Find(f[x]);}void Merge(int x, int y){ int fx = Find(x), fy = Find(y); if (fx!=fy) { f[fx] = fy; ans[fy] += ans[fx]; }}int main(){ int T; scanf("%d",&T); while (T--) { tot = 0; for (i=1; i<=1000; i++) ans[i] = 1; for (i=1; i<=1000; i++) f[i] = i; scanf("%d",&n); while (n--) { cin >> cc; if (cc=='P') { tot++; scanf("%lf %lf %lf %lf",&a[tot].s.x,&a[tot].s.y,&a[tot].e.x,&a[tot].e.y); for (i=1; i<=tot-1; i++) if (intersect(a[i],a[tot])) { Merge(tot,i); } } else { scanf("%d",&k); printf("%d\n",ans[Find(k)]); } } if (T) printf("\n"); } return 0;}
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