Codeforces Round #331 (Div. 2)C. Wilbur and Points(模拟+STL)
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题目链接
题意:就是给出一组点,然后问如果Xa>=Xb&&Ya>=Yb的话,那么a点的编号必须必b点大,同时,点的权值要满足给出的w序列。
解法:先判断给出的w,与点的w的是不是能完全重合,不能的话就是no
然后,我们就按照给出的w顺序找出点的顺序,然后判断一下即可
AAA:比赛时候貌似又读错了题意,,昨天状态太差。。。
#include<bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#define X first#define Y second#define cl(a,b) memset(a,b,sizeof(a))typedef pair<int,int> P;const int maxn=100005;const int inf=1<<27;const LL mod=1e9+7;int w[maxn],w_[maxn],ww[maxn];map<int,set<P> >mp;vector<P> v;int main(){ int n;scanf("%d",&n); for(int i=0;i<n;i++){ int x,y;scanf("%d%d",&x,&y); ww[i]=y-x; mp[y-x].insert(P(x,y)); } for(int i=0;i<n;i++){ scanf("%d",&w_[i]);w[i]=w_[i]; } sort(ww,ww+n); sort(w_,w_+n); for(int i=0;i<n;i++)if(w_[i]!=ww[i]){ puts("NO");return 0; } for(int i=0;i<n;i++){ set<P>::iterator it=mp[w[i]].begin(); v.pb(*it); mp[w[i]].erase(it); } // printf("v.size() = %d\n",v.size()); for(int i=1;i<n;i++){ if(v[i].X<v[i-1].X&&v[i].Y<=v[i-1].Y){ return puts("NO"); } } puts("YES"); for(int i=0;i<n;i++){ printf("%d %d\n",v[i].X,v[i].Y); } return 0;}
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