hdu 1558 Segment set

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思路: 计算几何+并查集
分析:
1 有n个操作，最后求有几个集合或者说是连通分量
2 对于输入一条线段我们就去前面找能够和它相交的线段，利用并查集进行合并并且更新rank数组，rank[x]数组保存的是以x为跟节点的集合的线段的数量
3 这一题难点就是线段的相交判断

代码:

 

#include<cmath>

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;



const double eps = 1e-8;

const double INF = 1<<30;

const int MAXN = 1010;



struct Node{

    double x1;

    double y1;

    double x2;

    double y2;

};

Node node[MAXN];



int n;

int father[MAXN];

int rank[MAXN];



void init(int m){

    for(int i = 0 ; i <= m ; i++){

        father[i] = i;

        rank[i] = 1;

    }

}



int find(int x){

    if(father[x] != x){

        int fa = father[x];

        father[x] = find(fa);

        rank[x] += rank[fa];

    }

    return father[x];

}



double multiply1(Node a , Node b){

    return (a.x1-a.x2)*(b.y1-a.y1)-(a.y1-a.y2)*(b.x1-a.x1);

}



double multiply2(Node a , Node b){

    return (a.x1-a.x2)*(b.y2-a.y1)-(a.y1-a.y2)*(b.x2-a.x1);

}



bool inter(Node a , Node b){

    if(max(a.x1,a.x2) >= min(b.x1,b.x2) &&

            max(b.x1,b.x2) >= min(a.x1,a.x2) &&

            max(a.y1,a.y2) >= min(b.y1,b.y2) &&

            max(b.y1,b.y2) >= min(a.y1,a.y2) &&

            multiply1(a,b) * multiply2(a,b) <= eps &&

            multiply1(b,a) * multiply2(b,a) <= eps)

        return true;

    else return false;

}

void solve(){

    for(int i = 1 ; i < n ; i++){

        if(inter(node[i] , node[n])){

            int fx = find(i); 

            int fy = find(n);

            if(fx != fy){

                father[fy] = fx;

                rank[fx] += rank[fy];

            }

        }          

    }

}



int main(){

    int cas , m , k;

    bool isFirst = true;

    char c;

    scanf("%d" , &cas);

    while(cas--){

        if(isFirst)

            isFirst = false;

        else

            puts("");

        n = 1;

        scanf("%d%*c" , &m); 

        init(m);

        while(m--){

            c = getchar(); 

            if(c == 'P'){

                scanf("%lf" ,&node[n].x1);       

                scanf("%lf" ,&node[n].y1);       

                scanf("%lf" ,&node[n].x2);       

                scanf("%lf%*c" ,&node[n].y2);       

                if(node[n].x1 > node[n].x2){

                    swap(node[n].x1 , node[n].x2); 

                    swap(node[n].y1 , node[n].y2); 

                }

                solve();

                n++;

            }

            else{

                scanf("%d%*c" , &k); 

                int fa = find(k);

                printf("%d\n" , rank[fa]);

            }

        }

    }

    return 0;

}