HDU4612 Warm up —— 边双联通分量 + 重边 + 缩点 + 树上最长路
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4612
Warm up
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7206 Accepted Submission(s): 1681
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.
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.
1 2
1 3
1 4
2 3
0 0
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e5+; struct Edge
{
int to, next;
}edge[MAXN*];
int tot, head[MAXN];
vector<int>g[MAXN]; void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
} int index, dfn[MAXN], low[MAXN];
int top, Stack[MAXN], instack[MAXN];
int block, belong[MAXN]; void Tarjan(int u, int pre)
{
dfn[u] = low[u] = ++index;
Stack[top++] = u;
instack[u] = true;
for(int i = head[u]; i!=-; i = edge[i].next)
{
//因为一对点之间可能有多条边,所以不能根据v是否为上一个点来防止边是否被重复访问。而需要根据边的编号
if((i^)==pre) continue;
int v = edge[i].to;
if(!dfn[v])
{
Tarjan(v, i);
low[u] = min(low[u], low[v]);
}
else if(instack[v])
low[u] = min(low[u], dfn[v]);
} if(low[u]==dfn[u])
{
block++;
int v;
do
{
v = Stack[--top];
instack[v] = false;
belong[v] = block;
}while(v!=u);
}
} int diameter, endpoint;
int dfs(int u, int pre, int dep)
{
if(dep>diameter) { endpoint = u; diameter = dep; }
for(int i = ; i<g[u].size(); i++)
if(g[u][i]!=pre)
dfs(g[u][i], u, dep+);
} void init(int n)
{
tot = ;
memset(head, -, sizeof(head)); index = ;
memset(dfn, , sizeof(dfn));
memset(low, , sizeof(low)); top = ;
memset(instack, false, sizeof(instack)); block = ;
for(int i = ; i<=n; i++)
belong[i] = i, g[i].clear();
} int main()
{
int n, m;
while(scanf("%d%d", &n, &m) && (n||m) )
{
init(n);
for(int i = ; i<=m; i++)
{
int u, v;
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
} Tarjan(, -);
for(int u = ; u<=n; u++)
for(int i = head[u]; i!=-; i = edge[i].next)
{
int v = edge[i].to;
if(belong[u]!=belong[v])
g[belong[u]].push_back(belong[v]);
} endpoint = , diameter = ;
dfs(, -, );
dfs(endpoint, -, );
printf("%d\n", block--diameter);
}
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e5+; struct Edge
{
int from, to, next;
}edge[MAXN*];
int tot, head[MAXN]; void addedge(int u, int v)
{
edge[tot].from = u;
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
} int index, dfn[MAXN], low[MAXN];
int top, Stack[MAXN], instack[MAXN];
int block, belong[MAXN]; void Tarjan(int u, int pre)
{
dfn[u] = low[u] = ++index;
Stack[top++] = u;
instack[u] = true;
for(int i = head[u]; i!=-; i = edge[i].next)
{
//因为一对点之间可能有多条边,所以不能根据v是否为上一个点来防止边是否被重复访问。而需要根据边的编号
if((i^)==pre) continue;
int v = edge[i].to;
if(!dfn[v])
{
Tarjan(v, i);
low[u] = min(low[u], low[v]);
}
else if(instack[v])
low[u] = min(low[u], dfn[v]);
} if(low[u]==dfn[u])
{
block++;
int v;
do
{
v = Stack[--top];
instack[v] = false;
belong[v] = block;
}while(v!=u);
}
} int diameter, endpoint;
int dfs(int u, int pre, int dep)
{
if(dep>diameter) { endpoint = u; diameter = dep; }
for(int i = head[u]; i!=-; i = edge[i].next)
if(edge[i].to!=pre)
dfs(edge[i].to, u, dep+);
} void init(int n)
{
tot = ;
memset(head, -, sizeof(head)); index = ;
memset(dfn, , sizeof(dfn));
memset(low, , sizeof(low)); top = ;
memset(instack, false, sizeof(instack)); block = ;
for(int i = ; i<=n; i++)
belong[i] = i;
} int main()
{
int n, m;
while(scanf("%d%d", &n, &m) && (n||m) )
{
init(n);
for(int i = ; i<=m; i++)
{
int u, v;
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
} Tarjan(, -);
tot = ;
memset(head, -, sizeof(head));
for(int i = ; i<*m; i++)
{
int u = edge[i].from, v = edge[i].to;
if(belong[u]!=belong[v])
addedge(belong[u], belong[v]);
} endpoint = , diameter = ;
dfs(, -, );
dfs(endpoint, -, );
printf("%d\n", block--diameter);
}
}
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