CodeForces 404C Restore Graph (构造)

题意：让人构造一个图，满足每个结点边的数目不超过 k，然后给出每个结点到某个结点的最短距离。

析：很容易看出来如果可能的话，树是一定满足条件的，只要从头开始构造这棵树就好，中途超了int。。。找了好久。

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 = 100000 + 10;
const int mod = 100000000;
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;
}

vector<int> G[maxn];
int main(){
  scanf("%d %d", &n, &m);
  int rt = -1, cnt = 0;
  int mmax = 0;
  for(int i = 1; i <= n; ++i){
    int x;
    scanf("%d", &x);
    G[x].push_back(i);
    if(x == 0)  rt = i;
    mmax = max(mmax, x);
  }

  if(G[1].size() > G[0].size() * m || G[0].size() != 1){
    printf("-1\n");
    return 0;
  }
  for(int i = 2; i <= mmax; ++i)
    if(G[i].size() > (LL)G[i-1].size() * (m-1)){
      printf("-1\n");
      return 0;
    }

  printf("%d\n", n-1);
  for(int i = 0; i < mmax; ++i){
    bool ok = true;
    int p = 0;
    for(int j = 0; j < G[i].size() && ok; ++j){
      int u = G[i][j];
      for(int k = (i != 0); k < m && ok; ++k){
        printf("%d %d\n", u, G[i+1][p++]); 
        if(p == G[i+1].size())  ok = false;
      }
    }
  }
  return 0;
}