https://codeforces.com/contest/1283/problem/D
题意:在一条无限长的坐标轴上,给你n颗树,m个人。求所有人到树的最短距离的总和和坐标。
解法:多源bfs,map标记。
//#include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 1000000007 #define PI acos(-1) using namespace std; typedef long long ll ; map<int , int>ma; struct node{ int x , st ; node(int x , int st):x(x),st(st){} }; int main() { queue<node>q; int n , m ; scanf("%d%d" , &n , &m); for(int i = 0 ; i < n ; i++) { int x ; scanf("%d" , &x); q.push(node(x , 0)); ma[x] = 1; } int ans = 0 ; ll sum = 0 ; vector<int>v; while(!q.empty()) { node a = q.front(); q.pop(); if(ma[a.x + 1] != 1) { ans++; q.push(node(a.x + 1 , a.st+1)); sum += a.st + 1 ; v.push_back(a.x + 1); ma[a.x+1] = 1 ; } if(ans == m) break ; if(ma[a.x - 1] != 1) { ans++; q.push(node(a.x - 1 , a.st+1)); sum += a.st + 1 ; v.push_back(a.x - 1); ma[a.x-1] = 1 ; } if(ans == m) break ; } cout << sum << endl ; for(auto i : v) { cout << i << " " ; } cout << endl ; return 0 ; }
原文地址:https://www.cnblogs.com/nonames/p/12237908.html
时间: 2024-10-08 11:00:13