Codeforces Round #331 (Div. 2)C. Wilbur and Points 贪心

C. Wilbur and Points

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/596/problem/C

Description

Wilbur is playing with a set of n points on the coordinate plane. All points have non-negative integer coordinates. Moreover, if some point (x, y) belongs to the set, then all points (x', y'), such that 0 ≤ x' ≤ x and 0 ≤ y' ≤ y also belong to this set.

Now Wilbur wants to number the points in the set he has, that is assign them distinct integer numbers from 1 to n. In order to make the numbering aesthetically pleasing, Wilbur imposes the condition that if some point (x, y) gets number i, then all (x',y') from the set, such that x' ≥ x and y' ≥ y must be assigned a number not less than i. For example, for a set of four points (0, 0), (0, 1), (1, 0) and (1, 1), there are two aesthetically pleasing numberings. One is 1, 2, 3, 4 and another one is 1, 3, 2, 4.

Wilbur's friend comes along and challenges Wilbur. For any point he defines it's special value as s(x, y) = y - x. Now he gives Wilbur some w1, w2,..., wn, and asks him to find an aesthetically pleasing numbering of the points in the set, such that the point that gets number i has it's special value equal to wi, that is s(xi, yi) = yi - xi = wi.

Now Wilbur asks you to help him with this challenge.

Input

The first line of the input consists of a single integer n (1 ≤ n ≤ 100 000) — the number of points in the set Wilbur is playing with.

Next follow n lines with points descriptions. Each line contains two integers x and y (0 ≤ x, y ≤ 100 000), that give one point in Wilbur's set. It's guaranteed that all points are distinct. Also, it is guaranteed that if some point (x, y) is present in the input, then all points (x', y'), such that 0 ≤ x' ≤ x and 0 ≤ y' ≤ y, are also present in the input.

The last line of the input contains n integers. The i-th of them is wi ( - 100 000 ≤ wi ≤ 100 000) — the required special value of the point that gets number i in any aesthetically pleasing numbering.

Output

If there exists an aesthetically pleasant numbering of points in the set, such that s(xi, yi) = yi - xi = wi, then print "YES" on the first line of the output. Otherwise, print "NO".

If a solution exists, proceed output with n lines. On the i-th of these lines print the point of the set that gets number i. If there are multiple solutions, print any of them.

Sample Input

5
2 0
0 0
1 0
1 1
0 1
0 -1 -2 1 0

Sample Output

YES
0 0
1 0
2 0
0 1
1 1

HINT

题意

给你一堆点，要求你找到一个集合，使得他的y[i]-x[i]=w[i]

且如果xj>=xi && yj>=yi，那么id[i]>id[j]

问你能否找到

题解：

我们贪心取最小就好了，然后再check一下就好了

check可以使用线段树，也可以先按照Y排序，然后再按照X排序，再扫一遍来check

代码

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;

struct node
{
    int x,y,z;
    int id;
};
node p[];
bool cmp(node a,node b)
{
    if(a.x==b.x&&a.y==b.y)return a.id<b.id;
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
int b[];
map<int,int> H;
vector<int> X;
vector<int> Y;
int tot = ;
vector<node> Q[];
int add[];
vector<node> Q2[];
int main()
{
    int n;scanf("%d",&n);
    for(int i=;i<=n;i++)
    {
        scanf("%d%d",&p[i].x,&p[i].y);
        p[i].z = p[i].y-p[i].x;
        if(H[p[i].z]==)
            H[p[i].z]=tot++;
        Q[H[p[i].z]].push_back(p[i]);
    }

    for(int i=;i<tot;i++)
        sort(Q[i].begin(),Q[i].end(),cmp);
    for(int i=;i<=n;i++)
    {
        scanf("%d",&b[i]);
        if(H[b[i]]==)
            return puts("NO");
        int t = H[b[i]];
        if(add[t]==Q[t].size())return puts("NO");
        X.push_back(Q[t][add[t]].x);
        Y.push_back(Q[t][add[t]].y);
        add[t]++;
    }
    for(int i=;i<n;i++)
    {
        node kkk;
        kkk.x = X[i],kkk.y = Y[i];
        kkk.id = i;
        Q2[kkk.y].push_back(kkk);
    }

    for(int i=;i<=;i++)
        sort(Q2[i].begin(),Q2[i].end(),cmp);
    for(int i=;i<=;i++)
    {
        if(Q2[i].size()<=)continue;
        for(int j=;j<Q2[i].size()-;j++)
        {
            if(Q2[i][j].id>Q2[i][j+].id)
                return puts("NO");
        }
    }
    puts("YES");
    for(int i=;i<X.size();i++)
    {
        printf("%d %d\n",X[i],Y[i]);
    }
}