https://vjudge.net/contest/389195#problem/F

InputThe input starts with an integerTT(1≤T≤1051≤T≤105), denoting the number of test cases.

For each test case, the first line contains two space-separated integersnnandkk(2≤n≤1052≤n≤105,0≤k≤1090≤k≤109), wherennis the number of days andkkis the fluctuation limit.

Theii-th line of the nextnnlines contains two space-separated integersliliandriri(0≤li≤ri≤1090≤li≤ri≤109), which is Cuber QQ's predicted interval in theii-th day. A prediction is believed to be correct if the stock priceii-th day lies betweenliliandriri, inclusive.

It is guaranteed that the sum of allnndoes not exceed106106.OutputFor each test case, first output a single lineYESorNO, that states whether Cuber QQ will win the 1 million price.

IfYES, in the next line, output a possible price series,a1,a2,…,ana1,a2,…,an, whereli≤ai≤rili≤ai≤ri(1≤i≤n1≤i≤n) and|ai−ai+1|≤k|ai−ai+1|≤k(1≤i≤n−11≤i≤n−1). The integers should be separated with space.
Sample Input

2
3 1
1 6
2 5
3 6
2 3
1 5
10 50

Sample Output

YES
2 3 3
NO

Sponsor

題意：每次能在一個能夠變換的區間變化，找區間

思路：

　　正反兩邊掃描, 把左邊的約束帶到右邊，反過來滿足後面的方法決定前面的改變．

　　同時滿足正反兩面變化的區間，最後剩下的為可行域（選最低的

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
#define debug(x) cout<<"debug:"<<#x<<" = "<<x<<endl;

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=1e6+10;
const int mod=1e9+7;

ll L[maxn] = {0};
ll R[maxn] = {0};
int main()
{
	ll l[maxn] = {0};
	ll r[maxn] = {0};
	ll n = 0 , k = 0;
	bool flag = 1;
	
    int T = 0;
    cin >> T;
    while(T--)
    {
        flag = 1;
        cin >> n >> k;
        for(int i = 0;i<n;i++)
        {
            cin >> L[i] >> R[i];
        }
        l[0] = L[0];
        r[0] = R[0];
        
        
        for(int i = 1;i<n;i++)
        {
            r[i] = min(R[i] , r[i-1] + k);
            l[i] = max(L[i] , l[i-1] - k);
            if(r[i] < l[i])
            {
            	flag = 0;
			}
        }
        for(int i = 1;i<n;i++)
        {
            r[n-1-i] = min(r[n-1-i] , r[n-i] + k);
            l[n-1-i] = max(l[n-1-i] , l[n-i] - k);
            if(r[i] < l[i])
			{
            	flag = 0;
			}
        }
        
        if(flag)
        {
            cout << "YES" <<endl;
            for(int i = 0;i<n;i++)
            {
          		cout << r[i];
            	if(i < n-1)
            	{
            		cout<<" ";
				}
			}
            cout <<endl;
        }
        else
        {
        	cout << "NO" <<endl;
		}
    }
    return 0;
}

/*
2
3 1
1 6
2 5
3 6
2 3
1 5
10 50
*/