标签：turn clu names 特殊 处理 space set show string
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)<<"\n"
#define DBN2(a,b) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define DBN3(a,b,c) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define DBN4(a,b,c,d) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define DBN5(a,b,c,d,e) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#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\n",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)<<"\n"
#define DBN2(a,b) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define DBN3(a,b,c) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define DBN4(a,b,c,d) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define DBN5(a,b,c,d,e) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#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\n",bcc_cnt-mx-1);
}
return 0;
}
标签：turn clu names 特殊 处理 space set show string
原文地址：https://www.cnblogs.com/cutemush/p/12682856.html
