Codeforces 141C Queue【思维】

C. Queue

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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
ai are correct.

Input
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.

Output
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.

Examples

Input

4
a 0
b 2
c 0
d 0

Output

a 150
c 170
d 180
b 160

Input

4
vasya 0
petya 1
manya 3
dunay 3

Output

-1

题目大意：

一共有N个人，给你每个人前边比这个人高的数量，让你找一个合法的安排，如果找不到输出-1.

思路：（为了方便描述，我们将h【i】=x设定为第i个人前边有x个人比他高）

1、首先我们按照h【i】从小到大排序，如果有h【i】>=i的情况发现，那么一定说明是无解的。

2、如果有解我们继续分析：

①对于第i个人来讲，其前边比他高的人数有h【i】个，那么前边比他矮的人数有i-1-h【i】个，那么我们可以将第i个人的高度设定为i-h【i】，保证前边一定有i-h【i】-1个比他矮的人。

②那么我们按照排好序的顺序设定第i个人的高度为i-h【i】，但是直接这样设定，会影响前边人的高度，那么对应将当前这个人前边的所有大于等于当前人高度的人的高度都加一即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
char s[15];
int x;
}a[3500];
int cmp(node a,node b)
{
return a.x<b.x;
}
int ans[3500];
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%s%d",a[i].s,&a[i].x);
}
sort(a+1,a+n+1,cmp);
int flag=0;
for(int i=1;i<=n;i++)
{
if(a[i].x>=i)flag=1;
ans[i]=i-a[i].x;
for(int j=1;j<i;j++)
{
if(ans[j]>=ans[i])
{
ans[j]++;
}
}
}
if(flag==1)
{
printf("-1\n");
continue;
}
for(int i=1;i<=n;i++)
{
printf("%s %d\n",a[i].s,ans[i]);
}
}
}