upc组队赛3 T-net【贪心】

T-net

题目描述

T-net which is a new telecommunications company, plans to install its base stations in the city. The places where base stations must be installed have been already specified. T-net has two types of antennas to be used in the base stations: (i)antennas with transmission radius a, and (ii) antennas with transmission radius b. Two antennas can communicate with each other if and only if both are inside the coverage area of each other. Antenna with smaller transmission radius of course is cheaper. T-net plans to minimize its cost while keeping the whole network connected. Precisely, T-net plans to
minimize its cost which is the sum of the transmission radii of all antennas. Interestingly, all base-station places are on a line. Help T-net construct a connected network with the minimum cost.

输入

The first line of the input contains three positive integers n, a and b (1 ⩽ n ⩽ 105 and 1 ⩽ a, b ⩽ 105 ) where n is the number of base stations, and a and b are radii as defined in the problem statement. The second line contains n distinct coordinates of base stations on the line with respect to an origin on the line. All coordinates are positive integers not more than 105 .

输出

If it is possible to construct a connected network, print the minimum cost in the output. Otherwise, print -1 .

样例输入

3 1 3
1 4 3

样例输出

7

提示

In the first case, there are four lamp pieces, two of each type. The worst possible lamp has value 1 + 1 = 2,
while the second worst possible lamp has value 2 + 1 = 3.

题解

一个相当复杂的贪心，参考[大神的题解]（https://blog.csdn.net/xiao__hei__hei/article/details/89066042）

代码

#include<bits/stdc++.h>
using namespace std;
#define memset(x,y) memset(x,y,sizeof(x))
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define print(x) printf("%d\n",x)
#define N 100001
typedef pair<int,int> P;
typedef long long ll;
const double eps=1e-8;
const double PI = acos(1.0);
const int inf = 0x3f3f3f3f;
const ll MOD = 998244353;
const int mod = 1e9+7;
const int maxn = 1e5 + 10;
ll ans ;
int flag ;
int n,a,b;
int pos[maxn];
int op[maxn];
int main()
{
  read3(n,a,b);
  if(a > b) swap(a,b);
  for(int i = 1; i <= n ;i++)
    read(pos[i]);
  sort(pos,pos + n + 1);
  pos[0] = pos[1] ; 
  pos[n + 1]  = pos[n];
  for(int i = 1 ; i <= n ; i++){
      if(op[i]) continue;
      if(pos[i] - pos[i - 1] > b && pos[i + 1] - pos[i] > b){
        printf("-1\n");
        return 0;
      }
      if(pos[i] - pos[i - 1] <= a && pos[i + 1] - pos[i] <=  a){
        op[i] = a;
        continue;
      }
      op[i] = b;
      while(op[i] == b){
        int j = i;
        while(j <= n){ 
          if(pos[i] + b < pos[j]) break;
          j++;
        }
        j--;
        flag = 0;
        for(int k = i + 1; k < j ; k++){
          if(pos[k] - pos[k - 1] > a || pos[k + 1] - pos[k] > a){
            flag = 1;
            break;
          }
        }
        if(!flag) break;
        op[j] = b;
        int l, r;
        for(l = i + 1 ; l < j; l++){
          if(pos[l] - pos[l - 1] <= a)
              op[l] = a;
          else
            break;
        }
        for(r = j - 1; r > i; r--){
          if(op[r]) break;
          if(pos[r + 1] - pos[r] <= a)
            op[r] = a;
          else
            break;
        }
        while(l <= r){
          op[l++] = b;
          while(l <= r && pos[l] - pos[l - 1] <= a){
            op[l++] = a ;
          }
        }
        i = j;
      }
  }
  ans = 0;
  for(int i = 1; i <= n; i++){
    ans += op[i];
  }
  printf("%lld\n",ans);
}