RevolC FaeLoN UVA

RevolC FaeLoN

 UVA - 10972 

题意：给一个无向图，现在要将所有的边变成有向，问至少还需要建几条边才能使得图强连通。

新的有向图强连通等价于原图边双连通。

/*************************************************************************
  > File Name: bb.cpp
  > Author: yijiull
  > Mail: 1147161372@qq.com 
  > Created Time: 2017年11月19日 星期三 21时10分33秒
 ************************************************************************/
#include <bits/stdc++.h>
using namespace std;
const int maxv = 1010;
const int maxe = 1000010;

struct Edge{
    int  u, v, nxt;
    int iscut;
    Edge(int u = 0, int v = 0, int nxt = 0, int iscut = 0) : u(u), v(v), nxt(nxt), iscut(iscut){}
}e[maxe<<1];
int cnt;
int head[maxv];
void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}
void add(int u, int v){
    e[cnt] = Edge(u, v, head[u]);
    head[u] = cnt++;
    e[cnt] = Edge(v, u, head[v]);
    head[v] = cnt++;
}

int pre[maxv], low[maxv], vis[maxv];
int dfsk, bcc_cnt;
int bccno[maxv];
void dfs(int u, int id){
    low[u] = pre[u] = ++dfsk;
    for(int i = head[u]; ~i; i = e[i].nxt){
        if(i == (id ^ 1)) continue;
        int v = e[i].v;
        if(!pre[v]){
            dfs(v, i);
            low[u] = min(low[v], low[u]);
            if(low[v] > pre[u]) e[i].iscut = e[i ^ 1].iscut = 1;
        }else low[u] = min(low[u], pre[v]);
    }
}
void dfs1(int u){
    vis[u] = 1;
    bccno[u] = bcc_cnt;
    for(int i = head[u]; ~i; i = e[i].nxt){
        if(e[i].iscut) continue;
        if(!vis[e[i].v]) dfs1(e[i].v);
    }
}
void find_bcc(int n){
    memset(vis, 0, sizeof(vis));
    memset(pre, 0, sizeof(pre));
    dfsk = bcc_cnt = 0;
    for(int i = 0; i < n; i++) if(!pre[i]) dfs(i, -1);
    for(int i = 0; i < n; i++) if(!vis[i]) bcc_cnt++, dfs1(i);
}

int deg[maxv];
int main(){
    int n, m;
    //freopen("in.txt", "r", stdin);
    while(scanf("%d %d", &n, &m) != EOF){
        init();
        int u, v;
        for(int i = 0; i < m; i++){
            scanf("%d %d", &u, &v);
            u--; v--;
            add(u, v);
        }
        find_bcc(n);
        memset(deg, 0, sizeof(deg));
        for(int u = 0; u < n; u++){
            for(int i = head[u]; ~i; i = e[i].nxt){
                int v = e[i].v;
                if(bccno[u] != bccno[v]) {
                    //deg[bccno[u]]++;
                    deg[bccno[v]]++;
                }
            }
        }
        if(bcc_cnt == 1) {
            puts("0");
            continue;
        }
        int a = 0, b = 0;
        for(int i = 1; i <= bcc_cnt; i++){
            if(deg[i] == 0) a++;
            if(deg[i] == 1) b++;
        }
        //cout<<a<<" "<<b<<endl;
        printf("%d
", a + (b + 1) / 2);
    }
}