CodeForces985G Team Players
2 seconds
256 megabytes
standard input
standard output
There are nn players numbered from 00 to n−1n−1 with ranks. The ii-th player has rank ii.
Players can form teams: the team should consist of three players and no pair of players in the team should have a conflict. The rank of the team is calculated using the following algorithm: let ii, jj, kk be the ranks of players in the team and i<j<ki<j<k, then the rank of the team is equal to A⋅i+B⋅j+C⋅kA⋅i+B⋅j+C⋅k.
You are given information about the pairs of players who have a conflict. Calculate the total sum of ranks over all possible valid teams modulo 264264.
The first line contains two space-separated integers nn and mm (3≤n≤2⋅1053≤n≤2⋅105, 0≤m≤2⋅1050≤m≤2⋅105) — the number of players and the number of conflicting pairs.
The second line contains three space-separated integers AA, BB and CC (1≤A,B,C≤1061≤A,B,C≤106) — coefficients for team rank calculation.
Each of the next mm lines contains two space-separated integers uiui and vivi (0≤ui,vi<n,ui≠vi0≤ui,vi<n,ui≠vi) — pair of conflicting players.
It's guaranteed that each unordered pair of players appears in the input file no more than once.
Print single integer — the total sum of ranks over all possible teams modulo 264264.
4 0
2 3 4
64
4 1
2 3 4
1 0
38
6 4
1 5 3
0 3
3 5
5 4
4 3
164
In the first example all 44 teams are valid, i.e. triples: {0, 1, 2}, {0, 1, 3}, {0, 2, 3} {1, 2, 3}.
In the second example teams are following: {0, 2, 3}, {1, 2, 3}.
In the third example teams are following: {0, 1, 2}, {0, 1, 4}, {0, 1, 5}, {0, 2, 4}, {0, 2, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}.
AC代码为:
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i = x ;i <= y; ++ i)
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
template<typename T>inline void read(T&x)
{
char c;int sign = 1;x = 0;
do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c));
do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c));
x *= sign;
}
const int N = 2e5 + 20;
ull a,b,c,n,m;
ull u[N],v[N],ans;
ull s1[N],s2[N],s3[N];
vector<int> g[N],f[N];
int main()
{
read(n); read(m);
read(a); read(b); read(c);
rep(i,1,m)
{
read(u[i]); read(v[i]);
if(u[i] > v[i]) swap(u[i],v[i]);
g[u[i]].push_back(v[i]);
f[v[i]].push_back(u[i]);
}
rep(i,0,n-1)
{
ull x = n - i - 1;
ans += a * i * (x * (x - 1) / 2);
ans += b * i * i * x;
ans += c * i * ((ull)i * (i - 1) / 2);
}
rep(i,1,m)
{
s1[ 0 ] += 1;
s1[u[i]] -= 1;
s1[ u[i] ] += n - u[i] - 2;
s1[u[i]+1] -= n - u[i] - 2;
s2[u[i]+1] += 1;
s2[ v[i] ] -= 1;
s2[ u[i] ] += u[i];
s2[u[i]+1] -= u[i];
s2[ v[i] ] += n - v[i] - 1;
s2[v[i]+1] -= n - v[i] - 1;
s3[v[i]+1] += 1;
s3[ n ] -= 1;
s3[ v[i] ] += v[i] - 1;
s3[v[i]+1] -= v[i] - 1;
}
rep(i,1,n)
s1[i] += s1[i - 1],
s2[i] += s2[i - 1],
s3[i] += s3[i - 1];
rep(i,0,n - 1)
{
ans -= a * i * s1[i];
ans -= b * i * s2[i];
ans -= c * i * s3[i];
}
rep(i,0,n-1) sort(g[i].begin(),g[i].end());
rep(i,0,n-1) sort(f[i].begin(),f[i].end());
rep(i,0,n-1)
{
int sz = g[i].size();
rep(j,0,sz - 1)
{
int k = j + 1;
while(k < sz)
{
ans += a * i;
ans += b * g[i][j];
ans += c * g[i][k];
k ++ ;
}
int SZ = g[g[i][j]].size();
rep(q,0,SZ - 1)
{
ans += a * i;
ans += b * g[i][j];
ans += c * g[g[i][j]][q];
}
}
sz = f[i].size();
rep(j,0,sz - 1)
{
int k = j + 1;
while(k < sz)
{
ans += a * f[i][j];
ans += b * f[i][k];
ans += c * i;
++ k;
}
}
}
rep(i,0,n-1)
{
int sz = g[i].size();
rep(j,0,sz - 1)
{
int t = j + 1,k = 0;
int SZ = g[g[i][j]].size();
while(t < sz && k < SZ)
{
if(g[i][t] == g[g[i][j]][k])
{
ans -= a * i;
ans -= b * g[i][j];
ans -= c * g[i][t];
++ t; ++ k;
}
else if(g[i][t] < g[g[i][j]][k]) ++ t;
else ++ k;
}
}
}
cout << ans << endl;
return 0;
}
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