HDU4612 Warm up

Posted 2021-03-01 cutemush

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4612

题意：给定一个无向图，问加一条边后最少的桥数是多少。

思路：找出边双连通分量后缩点成一棵树，然后我们要是加一条边使桥数最少，显然是去找树的直径，
所以两边DFS去找树的直径即可，注意这里很坑，重边是不算桥的，所以要特殊处理，下面给出两种实现的代码。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define Key_Value ch[ch[root][1]][0]
#define DBN1(a)           cerr<<#a<<"="<<(a)<<"
"
#define DBN2(a,b)         cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"
"
#define DBN3(a,b,c)       cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"
"
#define DBN4(a,b,c,d)     cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"
"
#define DBN5(a,b,c,d,e)   cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"
"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"
"
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=200000+5;
const int INF=0x3f3f3f3f;
const int P=1000000007;
const double PI=acos(-1.0);
template<typename T>
inline T read(T&x){
    x=0;int _f=0;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)_f|=(ch==‘-‘),ch=getchar();
    while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘,ch=getchar();
    return x=_f?-x:x;
}
struct Edge{
    int to;
    bool iscut;
    Edge(int _,bool __):to(_),iscut(__){}
};
int n,m,u,v,bcc_cnt,s,mx,dfs_clock,bccno[maxn],pre[maxn],low[maxn],dis[maxn];
vector<int>G[maxn];
vector<int>G2[maxn];
vector<Edge>edges;
void init(){
    bcc_cnt=dfs_clock=0;
    memset(pre,0,sizeof(pre));
    memset(bccno,0,sizeof(bccno));
    for (int i=1;i<=n;i++) G[i].clear(),G2[i].clear();
    edges.clear();
}
int dfs1(int u,int f){
    int lowu=pre[u]=++dfs_clock;
    for (int i=0;i<(int)G[u].size();i++){
        int v=edges[G[u][i]].to;
        if (!pre[v]){
            int lowv=dfs1(v,u);
            lowu=min(lowu,lowv);
            if (lowv>pre[u]) edges[G[u][i]].iscut=true,edges[G[u][i]^1].iscut=true;
        }
        else if (pre[v]<pre[u] && v!=f) lowu=min(lowu,pre[v]);
    }
    return low[u]=lowu;
}
void dfs2(int u){
    bccno[u]=bcc_cnt;
    for (int i=0;i<(int)G[u].size();i++){
        Edge &e=edges[G[u][i]];int v=e.to;
        if (!bccno[v] && !e.iscut) dfs2(v);//天然避开了重边的问题，因为有重边那么重边的iscut肯定是false肯定会被合并到边双连通分量里
    }
}
void find_bcc(){
    for (int i=1;i<=n;i++) if (!pre[i]) dfs1(i,-1);
    for (int i=1;i<=n;i++) if (!bccno[i]) bcc_cnt++,dfs2(i);
}
void dfs3(int u,int f,int dis){
    if (dis>mx){
        mx=dis;
        s=u;
    }
    for (int i=0;i<(int)G2[u].size();i++){
        int v=G2[u][i];
        if (v==f) continue;
        dfs3(v,u,dis+1);
    }
}
int main(){
    while (~scanf("%d%d",&n,&m)&&n+m){
        init();
        for (int i=1;i<=m;i++){
            read(u),read(v);
            edges.push_back(Edge(v,false));
            edges.push_back(Edge(u,false));
            int m=edges.size();
            G[u].push_back(m-2);
            G[v].push_back(m-1);
        }
        find_bcc();
        if (bcc_cnt==1){
            puts("0");
            continue;
        }
        for (int u=1;u<=n;u++){
            for (int i=0;i<(int)G[u].size();i++){
                int v=edges[G[u][i]].to;
                if (bccno[u]!=bccno[v]){
                    G2[bccno[u]].push_back(bccno[v]);
                }
            }
        }
        mx=0;
        dfs3(1,-1,0);
        dfs3(s,-1,0);
        printf("%d
",bcc_cnt-mx-1);
    }
    return 0;
}

　　

#pragma comment(linker, "/STACK:102400000,102400000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define Key_Value ch[ch[root][1]][0]
#define DBN1(a)           cerr<<#a<<"="<<(a)<<"
"
#define DBN2(a,b)         cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"
"
#define DBN3(a,b,c)       cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"
"
#define DBN4(a,b,c,d)     cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"
"
#define DBN5(a,b,c,d,e)   cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"
"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"
"
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=200000+5;
const int INF=0x3f3f3f3f;
const int P=1000000007;
const double PI=acos(-1.0);
template<typename T>
inline T read(T&x){
    x=0;int _f=0;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)_f|=(ch==‘-‘),ch=getchar();
    while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘,ch=getchar();
    return x=_f?-x:x;
}
int n,m,u,v,bcc_cnt,s,mx,dfs_clock,bccno[maxn],pre[maxn],low[maxn];
vector<int>G[maxn];
vector<int>G2[maxn];
stack<int>S;
void init(){
    bcc_cnt=dfs_clock=0;
    memset(pre,0,sizeof(pre));
    for (int i=1;i<=n;i++) G[i].clear(),G2[i].clear();
    while (!S.empty())S.pop();
}
int dfs1(int u,int f){
    int lowu=pre[u]=++dfs_clock,k=0;
    S.push(u);
    for (int i=0;i<(int)G[u].size();i++){
        int v=G[u][i];
        if (v==f && !k){//重边保证可以走
            k++;
            continue;
        }
        if (!pre[v]){
            int lowv=dfs1(v,u);
            lowu=min(lowu,lowv);
        }
        else if (pre[v]<pre[u]) lowu=min(lowu,pre[v]);
    }
    if (lowu==pre[u]){
        bcc_cnt++;
        for (;;){
            int v=S.top();S.pop();
            bccno[v]=bcc_cnt;
            if (v==u) break;
        }
    }
    return low[u]=lowu;
}
void find_bcc(){
    for (int i=1;i<=n;i++) if (!pre[i]) dfs1(i,-1);
}
void dfs2(int u,int f,int dis){
    if (dis>mx){
        mx=dis;
        s=u;
    }
    for (int i=0;i<(int)G2[u].size();i++){
        int v=G2[u][i];
        if (v==f) continue;
        dfs2(v,u,dis+1);
    }
}
int main(){
    while (~scanf("%d%d",&n,&m)&&n+m){
        init();
        for (int i=1;i<=m;i++){
            read(u),read(v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        find_bcc();
        if (bcc_cnt==1){
            puts("0");
            continue;
        }
        for (int u=1;u<=n;u++){
            for (int i=0;i<(int)G[u].size();i++){
                int v=G[u][i];
                if (bccno[u]!=bccno[v]){
                    G2[bccno[u]].push_back(bccno[v]);
                }
            }
        }
        mx=0;
        dfs2(1,-1,0);
        dfs2(s,-1,0);
        printf("%d
",bcc_cnt-mx-1);
    }
    return 0;
}