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发表于 更新于 本文字数： 650 阅读时长 ≈ 2 分钟

洛谷P3384，树剖板子题，树状数组维护区间操作

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#include <bits/stdc++.h>

using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

ll n,m,r,mod=1e9+7,cnt,v[(int)1e5+9],dep[(int)1e5+9],f[(int)1e5+9],sz[(int)1e5+9],son[(int)1e5+9],top[(int)1e5+9],id[(int)1e5+9];

vector<int> g[(int)1e5+9];

struct BIT
{
    ll a[(int)1e5 + 9],c[(int)1e5+9];
    int n;
    BIT(int x){
    	n=x;
    	for(int i=0;i<=n;++i)a[i]=c[i]=0;
    }
    void adda(int p,ll x)
    {
        for (; p <=n ; p += lowbit(p))
        {
            a[p] =(a[p]+x)%mod;
        }
    }
    void addc(int p,ll x)
    {
        for (; p <=n ; p += lowbit(p))
        {
            c[p] =(c[p]+x)%mod;
        }
    }

    void add(int l,int r,ll x){
    	adda(l,x);
    	adda(r+1,((-x)%mod+mod)%mod);
    	addc(l,(l-1)*x%mod);
    	addc(r+1,((r*(-x))%mod+mod)%mod);
    }

    ll qa(int p){
    	ll res=0;
    	for(;p;p-=lowbit(p))res=(res+a[p])%mod;
    	return res;
    }
    ll qc(int p){
    	ll res=0;
    	for(;p;p-=lowbit(p))res=(res+c[p])%mod;
    	return res;
    }

    ll query(int l,int r){
    	ll res=((qa(r)*r%mod-qc(r))%mod+mod)%mod;
    	res=((res-(qa(l-1)*(l-1)%mod-qc(l-1))%mod)%mod+mod)%mod;
    	return res;
    }
};

void dfs1(int u,int fa,int deep){
	dep[u]=deep,f[u]=fa,sz[u]=1;
	int ma=0;
	for(auto it:g[u]){
		if(it!=fa){
			dfs1(it,u,deep+1);
			sz[u]+=sz[it];
			if(ma<sz[it]){
				ma=sz[it],son[u]=it;
			}
		}
	}
}

void dfs2(int u,int fa,BIT &tr){
	top[u]=fa,id[u]=++cnt;
	if(v[u]){
		tr.add(id[u],id[u],v[u]);
	}
	if(!son[u])return ;
	dfs2(son[u],fa,tr);
	for(auto it:g[u]){
		if(it!=son[u]&&it!=f[u]){
			dfs2(it,it,tr);
		}
	}

}

void add(int u,int v,int x,BIT &tr){
	while(top[u]!=top[v]){
		if(dep[top[u]]<dep[top[v]])swap(u,v);
		tr.add(id[top[u]],id[u],x);
		u=f[top[u]];
	}
	if(dep[u]>dep[v])swap(u,v);
	tr.add(id[u],id[v],x);
}

ll query(int u,int v,BIT &tr){
	ll res=0;
	while(top[u]!=top[v]){
		if(dep[top[u]]<dep[top[v]])swap(u,v);
		res=(res+tr.query(id[top[u]],id[u]))%mod;
		u=f[top[u]];
	}
	if(dep[u]>dep[v])swap(u,v);
	res+=tr.query(id[u],id[v]);
	return res%mod;
}



int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>m>>r>>mod;
    BIT b(n);
    for(int i=1;i<=n;++i)cin>>v[i];
    for(int i=1,x,y;i<n;++i){
    	cin>>x>>y;
    	g[x].push_back(y),g[y].push_back(x);
    }
    dfs1(r,0,1);
    dfs2(r,r,b);
    for(int i=1,opt,x,y,z;i<=m;++i){
    	cin>>opt;
    	if(opt==1){
    		cin>>x>>y>>z;
    		add(x,y,z%mod,b);
    	}else if(opt==2){
    		cin>>x>>y;
    		cout<<query(x,y,b)<<"\n";
    	}else if(opt==3){
    		cin>>x>>z;
    		b.add(id[x],id[x]+sz[x]-1,z%mod);
    	}else {
    		cin>>x;
    		cout<<b.query(id[x],id[x]+sz[x]-1)<<"\n";
    	}
    }
    return 0;
}

发表于 更新于 本文字数： 252 阅读时长 ≈ 1 分钟

洛谷P3390

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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

const int mod=1e9+7;

ll n,k;

struct Matrix{
	ll a[109][109];
	friend Matrix operator*(const Matrix &x,const Matrix &y){
		Matrix res;
		for(int i=1;i<=n;++i)for(int j=1;j<=n;++j)res.a[i][j]=0;
		for(int i=1;i<=n;++i)for(int j=1;j<=n;++j)for(int k=1;k<=n;++k)res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
		return res;
	}
};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    Matrix a;
    Matrix res;
    f(i,1,n)f(j,1,n)cin>>a.a[i][j];
    f(i,1,n)res.a[i][i]=1;
    for(;k>0;k>>=1,a=a*a)if(k&1)res=res*a;
    f(i,1,n)f(j,1,n)cout<<res.a[i][j]<<" \n"[n==j];
    
    return 0;
}

发表于 更新于 本文字数： 34 阅读时长 ≈ 1 分钟

发表于 更新于 分类于 本文字数： 249 阅读时长 ≈ 1 分钟

1995

在此题中。

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#include <algorithm>
#include <bitset>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <cmath>
#include <stack>
#include <set>
#include <queue>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define sz(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

const int mod=1e9+7;

ll powmod(ll a,ll b){
    ll res=1;
    for(;b>0;b>>=1,a=a*a%mod)if(b&1)res=res*a%mod;
    return res;
}

ll n,k,s[(int)5e3+9][(int)5e3+9];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    s[0][0]=1;
    for(int i=1;i<=k;++i){
        for(int j=1;j<=k;++j){
            s[i][j]=(s[i-1][j-1]+s[i-1][j]*j)%mod;
        }
    }
    ll xjm=1,ans=0;
    for(int i=0;i<=k;++i){
        ans=(ans+s[k][i]*xjm%mod*powmod(2ll,n-i))%mod,xjm=xjm*(n-i)%mod;
    }
    
    cout<<ans<<"\n";
    return 0;
}

发表于 更新于 本文字数： 209 阅读时长 ≈ 1 分钟

右上角的元素重要性大于左下角的

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int tt, n;

struct dsu
{
    dsu(int n)
    {
        p.resize(n + 1);
        s.resize(n + 1);
        for (int i = 1; i <= n; i++)
            p[i] = i, s[i] = 1;
    }
    vector<int> p, s;
    int find(int x)
    {
        if (x != p[x])
            p[x] = find(p[x]);
        return p[x];
    }
    void merge(int x, int y)
    {
        int u = find(x), v = find(y);
        if (u != v)
        {
            p[v] = u;
            s[u] += s[v];
        }
    }
};

void solve()
{
    cin >> n;
    vector<vector<int>> a(n + 1, vector<int>(n + 1));
    dsu d(2 * n);
    for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>a[i][j];

    for(int i=1;i<=n;++i){
        for(int j=i+1;j<=n;++j){
            if(a[i][j]>a[j][i]){
                if(d.find(i)!=d.find(j)){
                    d.merge(i,j+n);
                    d.merge(j,i+n);
                    swap(a[i][j],a[j][i]);
                }
            }else if(a[i][j]<a[j][i]){
                if(d.find(i)!=d.find(j+n)){
                    d.merge(i,j);
                    d.merge(i+n,j+n);
                }else swap(a[i][j],a[j][i]);
            }
        }
    }

    f(i, 1, n) f(j, 1, n) cout << a[i][j] << " \n"[j == n];
}

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