查看CodeForces/397A的源代码 ←CodeForces/397A

[[分类:模拟与排序]][[分类:离散化]]
==摘要==
{{信息题|On Segment's Own Points|http://acm.hust.edu.cn/vjudge/contest/view.action?cid{{=}}70787#problem/G|1|100|数据类型|3|time=2015-02-28 13:28:11}}
*来自寒假练习：[http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70787 2015 Winter Day 5 div1] G题
*原题链接：http://codeforces.com/problemset/problem/397/A
==题意==
给出一堆区间，找出[l,r]内不属于任何区间的部分。
==题解==
离散化排序判断即可。注意每个点可能出现多次。
==代码==
{{折叠|397A.cpp代码已折叠
|<pre>#include<cstdio>
#include<iostream>
#include<string>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<bitset>
#include<set>
#include<stack>
#include<sstream>
#include<utility>
using namespace std;
//数据类型//
#define llu unsigned long long
#define lld long long
//定义默认类型//
typedef lld num;
#define dsi(n) num n;scanf("%lld",&n)
#define si(n) scanf("%lld",&n)
//其它//
#define reset(x) memset(x,0,sizeof(x))
#define ci const num&
#define sqr(x) ((x)*(x))
#define f(i,n) for(num i=1;i<=n;++i)
#define ff(i,r,n) for(num i=r;i<=n;++i)
#define fi(n) f(i,n)
#define f0(i,n) for(num i=0;i!=n;++i)
#define fd(i,n) for(num i=n;i>=1;--i)
#define foreach(i,s) for(typeof(s.begin()) i=s.begin();i!=s.end();++i)
#define rforeach(i,s) for(typeof(s.rbegin()) i=s.rbegin();i!=s.rend();++i)
#define iforeach(i,s) int idx=0;for(typeof(s.begin()) i=s.begin();i!=s.end();++i,++idx)
#define Vector2 pair<num,num>
#define vector2(x,y) make_pair(x,y)
#define x first
#define y second
multiset<Vector2> s;//fixed:multi.//
num sgNum,steOpen=0;
int main()
{
    dsi(n);
    dsi(x);dsi(y);
    s.insert(vector2(x,2));
    s.insert(vector2(y,2));
    fi(n-1)
    {
        dsi(l);dsi(r);
        s.insert(vector2(l,0));
        s.insert(vector2(r,1));
    }
    num ans=0;
    foreach(i,s)
    {
        typeof(i) j=i;
        ++j;
        if(j==s.end())
        {
            cout<<ans;
            return 0;
        }
        if(i->y==0)++sgNum;
        if(i->y==1)--sgNum;
        if(i->y==2&&steOpen)
        {
            cout<<ans;
            return 0;
        }
        if(i->y==2)
        {
            steOpen=1;
        }
        if(steOpen&&!sgNum)
        {
            ans+=j->x-i->x;
        }
    }
}
</pre>
|code397A}}