[ZJOI2011] 最小割 - 最小割树

最小割树裸题

建树后，以每个点为根跑DFS求出距离矩阵，然后暴力回答询问即可

#include <bits/stdc++.h>

using namespace std;
#define int long long
const int maxn=6e2;
const int maxm=4e4;
const int inf=1e13;

int n,m,q;

//for the target graph
vector <pair<int,int> > g[maxn];

void clear1() {
    for(int i=1;i<=n;i++) g[i].clear();
}

inline void add(int u,int v,int c) {
    g[u].push_back(make_pair(v,c));
}

int cnt,p[maxn],tmp[maxn],S[maxn];

//cnt stands for the first few Uni-blocks
//p[i] represents the number of the point on the i-th point
//tmp array is used to copy the sort p array
//S[u] indicates the Unicom block number where u is in

void clear2() {
    cnt=0;
    memset(p,0,sizeof p);
    memset(tmp,0,sizeof tmp);
    memset(S,0,sizeof S);
}

struct GHT {
    int s,t,maxFlow,cur[maxn];

    int edgeNum=-1,head[maxn],to[maxm<<1],nxt[maxm<<1];
    int w[maxm<<1],f[maxm<<1];

    GHT() {memset(head,-1,sizeof(head));}

    inline void add_edge(int u,int v,int c) {
        nxt[++edgeNum]=head[u];head[u]=edgeNum;
        to[edgeNum]=v;w[edgeNum]=c;
    }

    int dep[maxn],gap[maxn];

    inline void bfs() {
        memset(dep,0,sizeof(dep));memset(gap,0,sizeof(gap));
        dep[t]=gap[1]=1;queue<int> Q;Q.push(t);
        while(!Q.empty()) { int u=Q.front();Q.pop();
            for(int i=head[u];i!=-1;i=nxt[i]) if(!dep[to[i]])
                ++gap[dep[to[i]]=dep[u]+1],Q.push(to[i]);
        }
    }

    int dfs(int u,int lastFlow) {
        int used=0,minFlow=0;
        if(u==t) {maxFlow+=lastFlow;return lastFlow;}
        for(int &i=cur[u];i!=-1;i=nxt[i])
            if(f[i]&&dep[to[i]]+1==dep[u])
                if(minFlow=dfs(to[i],min(lastFlow-used,f[i])))
                {   f[i]-=minFlow;f[i^1]+=minFlow;
                    if((used+=minFlow)==lastFlow) return used;
                }
        if(!(--gap[dep[u]++])) dep[s]=n+1;
        ++gap[dep[u]];return used;
    }

    inline int ISAP(int x,int y) {
        for(register int i=0;i<=edgeNum;++i) f[i]=w[i];
        maxFlow=0;s=x;t=y;bfs();while(dep[s]<=n) {
            for(register int i=0;i<=n;++i) cur[i]=head[i];
            dfs(s,inf);
        }return maxFlow;
    }

    void dfs(int u) { S[u]=cnt;
        for(int i=head[u];i!=-1;i=nxt[i])
            if(f[i]&&S[to[i]]!=cnt) dfs(to[i]);
    }

    void build(int l,int r) {
        if(l>=r) return ;
        int x=p[l],y=p[l+1],cut=ISAP(x,y),L=l,R=r;
        ++cnt;dfs(x);add(x,y,cut);add(y,x,cut);
        for(register int i=l;i<=r;++i) tmp[S[p[i]]==cnt?L++:R--]=p[i];
        for(register int i=l;i<=r;++i) p[i]=tmp[i];
        build(l,L-1);build(R+1,r);
    }
};

int vis[maxn],f[maxn],ans[maxn][maxn];

void clear3() {
    memset(vis,0,sizeof vis);
    memset(f,0,sizeof f);
    memset(ans,0,sizeof ans);
}

void dfs(int p) {
    vis[p]=1;
    for(int i=0;i<g[p].size();i++) {
        if(vis[g[p][i].first]) continue;
        f[g[p][i].first]=min(f[p],g[p][i].second);
        dfs(g[p][i].first);
    }
}

signed main() {
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--) {
        GHT tree;
        cin>>n>>m;
        for(int i=1;i<=n;i++) p[i]=i; //remember this
        for(int i=1;i<=m;i++) {
            int t1,t2,t3;
            cin>>t1>>t2>>t3;
            tree.add_edge(t1,t2,t3);
            tree.add_edge(t2,t1,t3);
        }
        tree.build(1,n);
        for(int i=1;i<=n;i++) {
            memset(vis,0,sizeof vis);
            f[i]=inf;
            dfs(i);
            for(int j=1;j<=n;j++) {
                if(vis[j]) ans[i][j]=f[j];
                else ans[i][j]=0;
            }
        }
        cin>>q;
        for(int i=1;i<=q;i++) {
            int lim;
            cin>>lim;
            int tot=0;
            for(int j=1;j<=n;j++) {
                for(int k=1;k<j;k++) {
                    if(ans[j][k]<=lim) ++tot;
                }
            }
            cout<<tot<<endl;
        }
        cout<<endl;
        clear1();
        clear2();
        clear3();
    }
}

原文地址：https://www.cnblogs.com/mollnn/p/12272802.html