Submission #705392

#TimeUsernameProblemLanguageResultExecution timeMemory
705392alvingogoTug of War (BOI15_tug)C++14
100 / 100
2139 ms9208 KiB
#include <bits/stdc++.h> #pragma GCC optimize("Ofast") #define AquA cin.tie(0);ios_base::sync_with_stdio(0); #define fs first #define sc second #define p_q priority_queue using namespace std; /* ok, bitset!!! wait, is it possible that their's only one contestent that wants one kind of graph? ok, let's just eliminate these cases, because sum_deg = 8n and we have4n points to consider if we make a graph from people i to left l_i and right r_i. if we consider all the people and make a graph if someone's deg = 1 then we know that we should take it and its neighbor, and wow, that's a matching problem! first: find out if it has a perfect matching iterate all the nodes deg = 1 we know people's deg = 2 so that L and R can do this SAMITorz second: bitsetorz rings, choose the odd edges or the even edges, and do knapsack to find out if we can make the difference less then k i want to make it like "b |= b<<k" or else maybe the right side should have strength in [a,b]; we can find out how much we add to the right side given an array A, an int x, for all elements C, you can add C to x or subtract C to x, find out if you can make abs(x) less then or equal to k isn't it knapsack? 30000*30000*2*20/64 = 9e8 :EEEEE: CF 1e9/s la or greedy greedy? fine, since it's the only method i can figure out let's gooooooooooo :TLEthinking: */ bitset<1200007> bt; vector<vector<pair<int,int> > > g; int main(){ AquA; bt[0]=1; int n,k; cin >> n >> k; g.resize(4*n); vector<int> deg(4*n),vis(4*n); int sum=0; for(int i=0;i<2*n;i++){ int a,b,s; cin >> a >> b >> s; a--; b--; sum+=s; g[a+2*n].push_back({i,s}); g[i].push_back({a+2*n,s}); g[b+3*n].push_back({i,s}); g[i].push_back({b+3*n,s}); deg[i]=2; deg[a+2*n]++; deg[b+3*n]++; } queue<int> q; for(int i=2*n;i<4*n;i++){ if(deg[i]==1){ q.push(i); } if(deg[i]==0){ cout << "NO\n"; return 0; } } while(q.size()){ auto h=q.front(); q.pop(); int flag=0; for(auto y:g[h]){ if(!vis[y.fs]){ flag=1; vis[y.fs]=1; if(h>=3*n){ bt<<=y.sc; } for(auto z:g[y.fs]){ deg[z.fs]--; if(deg[z.fs]==1){ q.push(z.fs); } } } } if(!flag){ cout << "NO\n"; return 0; } } for(int i=0;i<2*n;i++){ if(!vis[i]){ //dfs time :happy_mention: int a=0,b=0; for(int x=i,cnt=0,e=-1;cnt==0 || x!=i;cnt=1){ vis[x]=1; for(auto f:g[x]){ if(f.fs==e){ continue; } if(f.fs>=3*n){ a+=f.sc; } for(auto y:g[f.fs]){ if(vis[y.fs] && (y.fs!=i || cnt==0)){ } else{ if(f.fs>=3*n){ b+=y.sc; } x=y.fs; e=f.fs; break; } } break; } } //cout << a << " " << b << '\n'; bt=(bt<<a)|(bt<<b); } } for(int i=0;i<=sum;i++){ if(bt[i]){ if(abs(i-(sum-i))<=k){ cout << "YES\n"; return 0; } } } cout << "NO\n"; return 0; }
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