Codeforces 556D - Case of Fugitive

Posted 2020-09-29 Wisdom+.+

556D - Case of Fugitive

思路：将桥长度放进二叉搜索树中（multiset），相邻两岛距离按上限排序，然后二分查找桥长度匹配并删除。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2e5+5;
struct node
{
    ll x;
    ll y;
    int id;
    bool  operator <(const node &a)const
    {
        return y<a.y;
    } 
}l[N],b[N];
struct Node
{
    ll x;
    int id;
    bool operator <(const Node &a)const
    {
        return x<a.x;
    }
}a[N];
int ans[N];
multiset<Node>ss; 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); 
    int n,m;
    cin>>n>>m;
    cin>>b[0].x>>b[0].y;
    for(int i=1;i<n;i++)
    {
        cin>>b[i].x>>b[i].y;
        l[i-1].x=b[i].x-b[i-1].y;
        l[i-1].y=b[i].y-b[i-1].x;
        l[i-1].id=i;
    }
    for(int i=0;i<m;i++)
    {
        cin>>a[i].x;
        a[i].id=i+1;
        ss.insert(a[i]); 
    }
    sort(l,l+n-1);
    for(int i=0;i<n-1;i++)
    {
        auto it=ss.lower_bound(Node{l[i].x,0});
        if(it==ss.end())
        {
            cout<<"No"<<endl;
            return 0; 
        } 
        if(it->x<=l[i].y)
        {
            ans[l[i].id]=it->id; 
            ss.erase(it);
        } 
        else 
        {
            cout<<"No"<<endl;
            return 0; 
        }
    }
    cout<<"Yes"<<endl;
    for(int i=1;i<n-1;i++)
    cout<<ans[i]<<" ";
    cout<<ans[n-1]<<endl;
    return 0;
}