BZOJ 1123 && Luogu P3469 [POI2008]BLO-Blockade 割点+乘法原理

想了半天式子。。。最后在邓大师的帮助下想出此题。。。。QWQ我还是太菜了

对于一个非割点，ans+=2*(n-1);

对于一个割点，ans+= 

　

#include<cstdio>
#include<iostream>
#define R register int
using namespace std;
const int N=100010,M=500010;
inline int g() {
    R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
    do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
}
int n,m,cnt=1,num;
int vr[M<<1],nxt[M<<1],fir[N],dfn[N],low[N],sz[N];
long long ans[N]; bool cut[N];
inline void add(int u,int v) {vr[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt;}
void tarjan(int u) {
    dfn[u]=low[u]=++num,sz[u]=1; R sum=0,flg=0;
    for(R i=fir[u];i;i=nxt[i]) { R v=vr[i];
        if(!dfn[v]) {
            tarjan(v); 
            sz[u]+=sz[v]; low[u]=min(low[u],low[v]);
            if(low[v]>=dfn[u]) {
                ++flg; ans[u]+=1ll*sz[v]*(n-sz[v]);
                sum+=sz[v]; if(u!=1||flg>1) cut[u]=true;
            }
        } else low[u]=min(low[u],dfn[v]);
    }
    if(cut[u]) ans[u]+=1ll*(n-sum-1)*(sum+1)+n-1;
    else ans[u]=2*(n-1);
}
signed main() {
    n=g(),m=g();
    for(R i=1,u,v;i<=m;++i) {
        u=g(),v=g();
        if(u!=v) add(u,v),add(v,u);
    }
    tarjan(1);
    for(R i=1;i<=n;++i) printf("%lld
",ans[i]);
}

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2019.04.12