CodeForces 141C Queue (构造)

题意：n 个人在排队，然后给出每个人的前面比他身高高的人的数量hi，让你给出一种排列，并给出一种解。

析：首先，hi 小的要在前面，所以先进行排序，然后第一个人的 h1 必须为0，我们可以令身高为 1，然后对于第 i 个人，前面1 ~ i-1 个人中有 hi 个人

比他高，那么就有 i-1-hi 个人不比他高，所以他的身高的最佳情况就是 i - hi。不过这样会覆盖前面那个身高和他相等的人，所以我们把 1 ~ i-1 中所有的

身高比他高的人都加上1，这样就消除了影响。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3000 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
  string name;
  int num;
  Node() { }
  Node(string s, int n) : name(s), num(n) { }
  bool operator < (const Node &p) const{
    return num < p.num;
  }
};
Node a[maxn];
const int det = 100000;

int main(){
  cin >> n;
  for(int i = 0; i < n; ++i)
    cin >> a[i].name >> a[i].num;

  sort(a, a + n);
  if(1 == n && a[0].num){ cout << "-1" << endl;  return 0; }
  a[0].num = 1;
  for(int i = 1; i < n; ++i){
    if(a[i].num > i){
      cout << "-1" << endl;
      return 0;
    }
    a[i].num = i - a[i].num + 1;
    for(int j = 0; j < i; ++j)
      if(a[j].num >= a[i].num)  ++a[j].num;
  }
  for(int i = 0; i < n; ++i)
    cout << a[i].name << " " << a[i].num + det << endl;
  return 0;
}