In the Main Berland Bank n people stand in a queue at the cashier, everyone knows his/her height hi, and the heights of the other people in the queue. Each of them keeps in mind number ai — how many people who are taller than him/her and stand in queue in front of him.
After a while the cashier has a lunch break and the people in the queue seat on the chairs in the waiting room in a random order.
When the lunch break was over, it turned out that nobody can remember the exact order of the people in the queue, but everyone remembers his number ai.
Your task is to restore the order in which the people stood in the queue if it is possible. There may be several acceptable orders, but you need to find any of them. Also, you need to print a possible set of numbers hi — the heights of people in the queue, so that the numbers aiare correct.
The first input line contains integer n — the number of people in the queue (1 ≤ n ≤ 3000). Then n lines contain descriptions of the people as "namei ai" (one description on one line), where namei is a non-empty string consisting of lowercase Latin letters whose length does not exceed 10 characters (the i-th person's name), ai is an integer (0 ≤ ai ≤ n - 1), that represents the number of people who are higher and stand in the queue in front of person i. It is guaranteed that all names are different.
If there's no acceptable order of the people in the queue, print the single line containing "-1" without the quotes. Otherwise, print in n lines the people as "namei hi", where hi is the integer from 1 to 109 (inclusive), the possible height of a man whose name is namei. Print the people in the order in which they stand in the queue, starting from the head of the queue and moving to its tail. Numbers hi are not necessarily unique.
4 a 0 b 2 c 0 d 0
a 150 c 170 d 180 b 160
4 vasya 0 petya 1 manya 3 dunay 3
-1
題意：告訴n個人每個人的前面有多少人比他高，要求求出所有人的可能高度
分析：這個就是一個腦洞題，想明白之後非常的簡單。首先，假設h[i]表示第i個人前面比他高的人的數量，所以前面比他矮或等於他的人的數量不就是i-1-h[i]麼，所以如果i-1-h[i]小於0，那麼不就不存在這個人的身高了，那麼不就出問題了？之後，我們就假設這個人的身高就是i-h[i]並且，所有人的身高不相等，那麼我們只要每次求出來這個人的身高之後再去把前面大於等於這個人的身高加一就好了，因爲所有人的身高都不相等了，小於這個人的身高的人數我們固定它，那麼剩下的不就都是大於這個人的身高了麼？
代碼：
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxed=3000+10;
struct Node
{
string str;
int val,h;
}node[maxed];
int n;
int main()
{
bool cmp(Node,Node);
scanf("%d",&n);
for(int i=1;i<=n;i++)
cin>>node[i].str>>node[i].val;
sort(node+1,node+1+n,cmp);
for(int i=1;i<=n;i++){
if(i-1-node[i].val<0)
return printf("-1")*0;
node[i].h=i-node[i].val;
for(int j=1;j<i;j++)
if(node[j].h>=node[i].h)
node[j].h++;
}
for(int i=1;i<=n;i++)
cout<<node[i].str<<" "<<node[i].h<<endl;
}
bool cmp(Node n1,Node n2)
{
return n1.val<n2.val;
}