CodeForces 658C Bear and Forgotten Tree 3 (构造)

题意：构造出一个 n 个结点，直径为 m，高度为 h 的树。

析：先构造高度，然后再构造直径，都全了，多余的边放到叶子上，注意直径为1的情况。

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 = 100 + 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;
}

vector<P> ans;

int main(){
  int h;
  scanf("%d %d %d", &n, &m, &h);
  bool ok = true;
  int cnt = 1, last = 0;
  for(int i = 0; i < h; ++i){
    last = cnt;
    ans.push_back(P(cnt, cnt+1));
    ++cnt;
    if(cnt > n)  ok = false;
  }
  int idx = 0;
  if(m > h){  ans.push_back(P(1, cnt+1));  ++cnt; ++idx;  }
  if(cnt > n) ok = false;
  for(int i = 1; i < m-h; ++i){
    ans.push_back(P(cnt, cnt+1));
    ++cnt;
    ++idx;
    if(cnt > n)  ok = false;
  }
  while(cnt < n && m > 1)  ans.push_back(P(last, cnt+1)), ++cnt;
  if(idx > h || cnt != n) ok = false;
  if(!ok){ printf("-1\n");  return 0; }
  for(int i = 0; i < ans.size(); ++i)
    printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}