Full_of_Boys训练7总结

题目来源：2016沈阳区域赛

C.Recursive sequence

矩阵快速幂，思路来自oldz

#include <bits/stdc++.h>
typedef unsigned long long ll;
const ll mod=2147493647;
using namespace std;
ll A[7][7]={
    {1,0,0,0,0,0,0},
    {1,1,0,0,0,0,0},
    {1,2,1,0,0,0,0},
    {1,3,3,1,0,0,0},
    {1,4,6,4,1,0,0},
    {0,0,0,0,0,0,1},
    {0,0,0,0,1,2,1}};
ll tmp[7][7],c[7][7],d[7][7],ans[7][7],B[7][7];
inline void mul(ll a[][7], ll b[][7], int n){
    for(int i=0;i<n;++i)
        for(int j=0;j<n;++j)tmp[i][j]=0,c[i][j]=a[i][j],d[i][j]=b[i][j];
    for(int i=0;i<n;++i)
    for(int k=0;k<n;++k)if(c[i][k])
    for(int j=0;j<n;++j)if(d[k][j]){
        tmp[i][j] = (tmp[i][j]+(c[i][k]*d[k][j])%mod)%mod;
    }
    for(int i=0;i<n;++i)
        for(int j=0;j<n;++j)a[i][j]=tmp[i][j];
}
void mx_pow(ll ans[][7], ll A[][7], ll b){
    while(b){
        if(b&1)mul(ans,A,7);
        mul(A,A,7);
        b>>=1;
    }
}
ll n,a,b;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld%lld",&n,&a,&b);
        for(int i=0;i<7;++i)
            for(int j=0;j<7;++j)
                if(i!=j)ans[i][j]=0;
                else ans[i][j]=1;
        for(int i=0;i<7;++i)
            for(int j=0;j<7;++j) B[i][j]=A[i][j];
        mx_pow(ans,B,n-2);
        a%=mod;b%=mod;
        ll  res = (ans[6][0]%mod+(3*ans[6][1])%mod+(9*ans[6][2])%mod+(27*ans[6][3])%mod+(81*ans[6][4])%mod+(a*ans[6][5])%mod+(b*ans[6][6])%mod)%mod;
        printf("%lld\n",res);
    }
    return 0;
}

E.Counting Cliques

搜索，建图时有个比较重要的优化，写完代码会发现用到的边，只有由编号小到大的单向边。。。根本没想到搜。。。总结一下吧

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;
int n,m,s,x,y,tmp[111],cc,ans,mp[111][111],vis[111];
vector<int> G[111];
inline int ck(int x){
    for(int i=1;i<=cc;++i) if(mp[tmp[i]][x]==0) return 0;
    return 1;
}
inline void dfs(int u){
    if(cc==s){
        ++ans;
        return;
    }
    for(int i=0;i<G[u].size();++i){
        int v=G[u][i];
        if(!vis[v]&&v>u&&ck(v)){
            tmp[++cc]=v;
            vis[v]=1;
            dfs(v);
            --cc;
            vis[v]=0;
        }
    }
}
vector<int> tt;
inline void del(int x){
    for(int i=0;i<G[x].size();++i){
        int v=G[x][i];tt.clear();
        mp[x][v]=mp[v][x]=0;
        for(int j=0;j<G[v].size();++j){
            if(G[v][j]!=x)tt.push_back(G[v][j]);
        }
        G[v]=tt;
    }
    G[x].clear();
}
int T;
int main() {
    scanf("%d",&T);
    while(T--){
        ans=0;
        scanf("%d%d%d",&n,&m,&s);
        for(int i=1;i<=n;++i){
            G[i].clear();
            for(int j=1;j<=i;++j)mp[i][j]=mp[j][i]=0;
        }
        for(int i=1;i<=m;++i){
            scanf("%d%d",&x,&y);
            if(x>y)swap(x,y);
            G[x].push_back(y);//!!!!important!!!!
            mp[x][y]=mp[y][x]=1;
        }
        for(int i=1;i<=n;++i){
            cc=0;
            vis[i]=1;
            tmp[++cc]=i;
            dfs(i);
            vis[i]=0;
            del(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}

G.Do not pour out

二分+微积分，膜一发wst