HDU 4612 Warm up (树的直径 + 双联通)

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

题意：有N个点，M条边（有重边）的无向图，这样图中会可能有桥，问加一条边后，使桥最少，求该桥的数量

思路：缩点后，求出图中的桥的个数，然后重建图必为树，求出树的最长直径，在该直径的两端点连一边，则图中的桥会最少

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int maxn = 200100;
const int maxm = 2001000;

int head[maxn], head1[maxn], cnt, ccnt;
int n, m, dis[maxn];
int low[maxn], dfn[maxn], sta[maxn], bel[maxn];
bool iscut[maxn];
int top, idx, block, bri;

struct edge
{
int from, to, nxt;
bool cut;
} e[maxm];

struct edge1
{
int from, to, nxt;
} e1[maxm];

void init()
{
cnt = ccnt = 0;
idx = top = block = 0;
memset(dfn, 0, sizeof(dfn));
memset(iscut, 0, sizeof(iscut));
memset(bel, -1, sizeof(bel));
memset(head, -1, sizeof(head));
memset(head1, -1, sizeof(head1));
}

void add(int u, int v)
{
e[cnt].from = u;
e[cnt].to = v;
e[cnt].nxt = head[u];
e[cnt].cut = false;
head[u] = cnt++;
}

void add1(int u, int v)
{
e1[ccnt].from = u;
e1[ccnt].to = v;
e1[ccnt].nxt = head1[u];
head1[u] = ccnt++;
}

void dfs(int u)
{
for(int i = head1[u]; ~i; i = e1[i].nxt)
{
int v = e1[i].to;
if(dis[v] == -1)
{
dis[v] = dis[u] + 1;
dfs(v);
}
}
}

int bfs(int s)
{
memset(dis, -1, sizeof(dis));
dis[s] = 0;
queue <int> que;
que.push(s);
int Max = -1;
int ans = 0;
while(!que.empty())
{
int u = que.front();
que.pop();
for(int i = head1[u]; ~i; i = e1[i].nxt)
{
int v = e1[i].to;
if(dis[v] == -1)
{
dis[v] = dis[u] + 1;
if(dis[v] > Max)
{
Max = dis[v];
ans = v;
}
que.push(v);
}
}
}
return ans;
}

void tarjan(int u, int pre)
{
int v;
low[u] = dfn[u] = ++idx;
sta[++top] = u;
int child = 0, flag = 1;
for(int i = head[u]; ~i; i = e[i].nxt)
{
v = e[i].to;
if(flag && v == pre)
{
flag = 0;
continue;
}
if(!dfn[v])
{
child++;
tarjan(v, u);
low[u] = min(low[u], low[v]);
if(low[v] >= dfn[u])
{
iscut[u] = true;
if(low[v] > dfn[u])
{
e[i].cut = true;
e[i ^ 1].cut = true;
}
}
}
else
low[u] = min(low[u], dfn[v]);
}
if(child == 1 && pre < 0)
iscut[u] = false;
if(low[u] == dfn[u])
{
block++;
do
{
bel[sta[top]] = block;
}
while(sta[top--] != u);
}
}

int main()
{
while(~scanf("%d%d", &n, &m) && n + m)
{
init();
for(int i = 0; i < m; i++)
{
int u, v;
scanf("%d%d", &u, &v);
add(u, v);
add(v, u);
}
for(int i = 1; i <= n; i++)
{
if(!dfn[i])
tarjan(i, -1);
}
int ans = 0;
int N = 0;
for(int i = 0; i < cnt; i += 2)
{
int u = bel[e[i].from];
int v = bel[e[i].to];
N = max(N, max(u, v));
if(u != v)
{
add1(u, v);
add1(v, u);
}
ans += e[i].cut;
}
int len = 0;
int u, v;
memset(dis, -1, sizeof(dis));
dis[1] = 0;
dfs(1);
for(int i = 1; i <= N; i++)
{
if(dis[i] > len)
{
len = dis[i];
u = i;
}
}
memset(dis, -1, sizeof(dis));
dis[u] = 0;
dfs(u);
for(int i = 1; i <= N; i++)
{
if(dis[i] > len)
len = dis[i];
}
u = bfs(1);
v = bfs(u);
printf("%d\n", ans - len);
}
return 0;
}