Warm up
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3790 Accepted Submission(s): 862
Problem Description
N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
Note that there could be more than one channel between two planets.
Input
The input contains multiple cases.
Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
(2<=N<=200000, 1<=M<=1000000)
Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
A line with two integers '0' terminates the input.
Output
For each case, output the minimal number of bridges after building a new channel in a line.
Sample Input
4 4
1 2
1 3
1 4
2 3
0 0
Sample Output
0
Source
2013 Multi-University Training Contest 2
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先 tarjan然后缩点得到一颗树，求树的直径就好了
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 200010;
const int inf = 0x3f3f3f3f;
const int M = 1000010;
struct node
{
int next;
int to;
int id;
}edge[M], edge2[M];
bool vis[N];
bool flag[N];
int dist[N];
int head[N];
int head2[N];
int low[N];
int block[N];
int DFN[N];
bool instack[N];
stack<int>qu;
int br_cnt, n, m, end_p;
int tot, tot2, maxs, index, top, sccnum;
void addedge(int from, int to, int id)
{
edge[tot].id = id;
edge[tot].to = to;
edge[tot].next = head[from];
head[from] = tot++;
}
void addedge2(int from, int to)
{
edge2[tot].to = to;
edge2[tot].next = head2[from];
head2[from] = tot++;
}
void init()
{
index = 0;
sccnum = 0;
top = 0;
tot = 0;
tot2 = 0;
maxs = 0;
br_cnt = 0;
end_p = 0;
memset ( head, -1, sizeof(head) );
memset (head2, -1, sizeof(head2) );
memset ( low, 0, sizeof(low) );
memset (DFN, 0, sizeof(DFN) );
memset ( instack, 0, sizeof(instack) );
memset ( flag, 0, sizeof(flag) );
memset (vis, 0, sizeof(vis) );
}
void build()
{
int u, v;
while (!qu.empty() )
{
qu.pop();
}
for (int i = 1; i <= m; ++i)
{
scanf("%d%d", &u, &v);
addedge(u, v, i);
addedge(v, u, i);
}
}
void tarjan(int u, int fa)
{
qu.push(u);
instack[u] = 1;
DFN[u] = low[u] = ++index;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (fa == edge[i].id)
{
continue;
}
if ( !instack[v] )
{
tarjan(v, edge[i].id);
low[u] = min(low[u], low[v]);
if (DFN[u] < low[v])
{
br_cnt++;
}
}
else
{
low[u] = min(DFN[v], low[u]);
}
}
if (DFN[u] == low[u])
{
sccnum++;
int v;
while(1)
{
v = qu.top();
qu.pop();
instack[v] = 0;
block[v] = sccnum;
if(v == u)
{
break;
}
}
}
}
void bfs(int s)
{
queue<int>qu;
while ( !qu.empty() )
{
qu.pop();
}
memset ( vis, 0, sizeof(vis) );
qu.push(s);
dist[s] = 0;
vis[s] = 1;
maxs = 0;
while ( !qu.empty() )
{
int u = qu.front();
qu.pop();
for (int i = head2[u]; i != -1; i = edge2[i].next)
{
int v = edge2[i].to;
if (!vis[v])
{
vis[v] = 1;
dist[v] = dist[u] + 1;
qu.push(v);
if (maxs < dist[v])
{
maxs = dist[v];
end_p = v;
}
}
}
}
}
void solve()
{
for (int i = 1; i <= n; ++i)
{
if (DFN[i] == 0)
{
tarjan(i, -1);
}
}
for (int u = 1; u <= n; ++u)
{
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (block[u] != block[v])
{
addedge2(block[u], block[v]);
}
}
}
bfs(1);
bfs(end_p);
printf("%d\n", br_cnt - maxs);
}
int main()
{
while (~scanf("%d%d", &n, &m))
{
if (!n && !m)
{
break;
}
init();
build();
solve();
}
return 0;
}
