Luogu P4255 公主の#18文明游戏 题解

#include<bits/stdc++.h>
#define LL long long
#define mod 19260817
using namespace std;
inline int read(){
    int res=0,f=1;char ch=getchar();
    while(ch<'0'ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x<10) putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
int n,m,q,fa[400010],vis[400010],A,C,ans[400010],Z[400010];
LL fac[400010*4];
inline LL fpow(LL a,LL b){// 快速幂，求逆元
    LL s=1;
    while(b){
        if(b&1) s*=a,s%=mod;
        a*=a;a%=mod;
        b>>=1;
    }
    return s;
}
struct Edge{
    int x,y;
}E[400010];
struct Question{
    int op,x,y,z;
}Q[400010];
map<int,int> v[400010];
vector<int> g[400010];
inline int getfa(int x){// 并查集启发式合并
    int a=x,y;
    while(x!=fa[x]) x=fa[x];
    while(a!=x) y=a,a=fa[a],fa[y]=x; 
    return x;
}
inline void merge(int x,int y){
    x=getfa(x);y=getfa(y);
    if(x==y) return ;
    if(v[x].size()<v[y].size()) swap(x,y);// 选择小的合并到大的
    fa[y]=x;Z[x]+=Z[y];
    for(int i:g[y]){// 暴力遍历合并
        if(v[x][i]==0&&v[y][i]!=0) g[x].push_back(i);
        v[x][i]+=v[y][i];
    }
    Z[y]=0;// 清空
    v[y].clear();
    g[y].clear();
}
int main(){
    n=read(),m=read(),q=read(); 
    fac[0]=1;
    for(int i=1;i<=4*400000;i++){
        fac[i]=(LL)fac[i-1]*i%mod;// 预处理阶乘
    }
    for(int x,y,i=1;i<=n;i++){
        x=read();y=read();
        fa[i]=i;
        v[i].clear();
        g[i].clear();
        g[i].push_back(y);
        v[i][y]+=x;
        Z[i]=x;
    }
    for(int i=1;i<=m;i++){
        E[i].x=read(),E[i].y=read();
    }
    for(int i=1;i<=q;i++){// 先记录再倒序做
        Q[i].op=read();
        if(Q[i].op==1){
            Q[i].x=read(),Q[i].y=read(),Q[i].z=read();
            if(v[Q[i].x][Q[i].z]==0) g[Q[i].x].push_back(Q[i].z);
            v[Q[i].x][Q[i].z]+=Q[i].y;
            Z[Q[i].x]+=Q[i].y;
        }else if(Q[i].op==2){
            Q[i].x=read();
            vis[Q[i].x]=1;
        }else if(Q[i].op==3){
            Q[i].x=read(),Q[i].y=read(),Q[i].z=read();
        }
    }
    for(int i=1;i<=m;i++){// 先连不需要删除的边
        if(vis[i]==0) merge(E[i].x,E[i].y);
    }
    for(int i=q;i>=1;i--){
        if(Q[i].op==1){
            Q[i].x=getfa(Q[i].x);
            v[Q[i].x][Q[i].z]-=Q[i].y;// 减值
            Z[Q[i].x]-=Q[i].y;
            if(v[Q[i].x][Q[i].z]==0) g[Q[i].x].erase(find(g[Q[i].x].begin(),g[Q[i].x].end(),Q[i].z));// 如果这删完以后是 0 记得删 vector，否则会重复合并
        }else if(Q[i].op==2){
            merge(E[Q[i].x].x,E[Q[i].x].y);// 连边
        }else if(Q[i].op==3){
            Q[i].x=getfa(Q[i].x);
            A=C=0;A=Z[Q[i].x];C=v[Q[i].x][Q[i].z];
            LL tmp1=fac[C]*fac[A-Q[i].y]%mod;
            LL tmp2=fac[A]*fac[C-Q[i].y]%mod;
            tmp2=fpow(tmp2,mod-2);// 求逆元
            LL res=tmp1*tmp2;
            res+=mod;res%=mod;
            ans[i]=res;
        }
    }
    for(int i=1;i<=q;i++)
        if(Q[i].op==3) write(ans[i]),putchar('\n');
}