코딩/BOJ & 알고스팟
[BOJ 1939] 중량제한
hoon222y
2017. 8. 31. 12:30
https://www.acmicpc.net/problem/1939
주어진 도로의 중량제한을 넘어서지 않는 배달가능한 최대물품중량을 계산하는문제이다.
BFS와 이분탐색을 이용해서 문제를 해결하면 된다.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 | #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <stack> #include <deque> #include <queue> #include <cmath> #include <stdio.h> #define INF 1e9 typedef long long ll; using namespace std; int n,m,mid,dist[111111],s,e; bool visit[111111]; vector<vector<pair<int,int>>> v; bool solve(){ queue<int> q; q.push(s); while(!q.empty()){ int t = q.front(); if(t == e){ return true; } q.pop(); if(visit[t])continue; visit[t] = true; for(int i=0;i<v[t].size();i++){ int cost = v[t][i].second; int next = v[t][i].first; if(cost >= mid){ q.push(next); } } } return false; } int main(){ cin >> n >>m; v.resize(n+1); for(int i=1;i<=m;i++){ int a,b,c; cin >> a >> b >>c; v[a].push_back({b,c}); v[b].push_back({a,c}); } cin >>s >>e; int l = 1; int r = 10000000; int ans =0; while(l<=r){ memset(visit,false,sizeof(visit)); mid = (r+l)/2; cout << l << " " << r << " " <<mid; if(solve()){ cout << "can" <<endl; ans = mid; l = mid+1; }else{ cout << "false" <<endl; r = mid-1; } } cout << ans <<endl; return 0; } | cs |