区间重合判断

题目大意：

输入两个表示区间范围的整数[x,y]

然后输入N个无序区间[x1,y1], [x2, y2], [x3, y3]...

求解第一次输入的区间是否在N个无序区间组合成的大区间中。

 

法一：使用并查集，对每个区间合并到一个子树上，最后判断源区间的x和y的根是否相同。

#include<iostream>
using namespace std;

const int size = 100;
int father[size];
int rank[size];

void make_set(int n)
{
    for(int i = 1; i <= n; i ++){
        father[i] = i;    
        rank[i] = 1;
    }    
}

int find_set(int x)
{
    if(x != father[x]){
        father[x] = find_set(father[x]);    
    }    
    return father[x];
}

void Union(int x, int y)
{
    x = find_set(x);    
    y = find_set(y);
    if(x == y){
        return ;    
    }
    if(rank[x] < rank[y]){
        father[x] = y;    
    }
    else{
        father[y] = x;
        if(rank[x] == rank[y]){
            rank[x] ++;    
        }    
    }
}

int main()
{
    int x1, y1;
    cin >> x1 >> y1;
    int x, y;
    int n;
    cin >> n;
    make_set(size);
    while(n --){
        cin >> x >> y;
        if(x > y){
            swap(x, y);    
        }
        for(int i = x + 1; i <= y; i ++){
            Union(x, i);
        }
    }    
    if(find_set(x1) == find_set(y1)){
        cout << "yes" << endl;    
    }
    else{
        cout << "no" << endl;    
    }
    system("pause");
}

法二：将无序的目标区间排序后，再合并成几个有序的区间，然后把源区间和有序的目标区间比较。

#include<iostream>
#include<vector>
using namespace std;

typedef pair<int, int> section;

bool cmp(section a, section b)
{
    return a.first < b.first;    
}

int main()
{
    section src, tmp;
    cin >> src.first >> src.second;
    
    vector<section> v;
    while(cin >> tmp.first >> tmp.second, tmp.first | tmp.second){
        v.push_back(tmp);    
    }
    
    sort(v.begin(), v.end(), cmp);
    
    vector<section> res;
    vector<section>::iterator it = v.begin();
    int begin = it->first;
    for(; (it + 1) != v.end(); it ++){
        if(it->second < (it + 1)->first){
            res.push_back(make_pair(begin, it->second));    
            begin = (it + 1)->first;
        }    
    }
    
    bool solve = false;
    it = res.begin();
    for(; it != res.end(); it ++){
        if(src.first >= it->first && src.second <= it->second){
            solve = true;    
            break;
        }    
    }
    if(solve){
        cout << "in" << endl;    
    }
    else{
        cout << "out" << endl;    
    }
    system("pause");
}