# | TimeUTC-0 | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
94067 | updown1 | Tug of War (BOI15_tug) | C++17 | 173 ms | 10232 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
/*
nodes are spots (0-N-1:left, N - 2*N-1: right)
edges are people
weight of edge is strength
if there is only one edge in a node, remove and repeat
we are not left with cycles so all edges should have exactly two edges in it
find the possibility for each cycle
use bitset for quick dp
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define For(i, a, b) for(int i=a; i<b; i++)
#define ffi For(i, 0, 2*N)
#define ffj For(j, 0, M)
#define ffa ffi ffj
#define s <<" "<<
#define c <<" : "<<
#define w cout
#define e "\n"
#define pb push_back
#define mp make_pair
#define a first
#define b second
//#define int ll
const int MAXN=60000, INF=1000000000;
///500,000,000
int N, K, s1, s2, tot;
bool vis[MAXN];
vector<int> vals;
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