2015 ACM Amman Collegiate Programming Contest H.Bridges【边双联通+求树上最长链(树的直径)】
2017-08-11 15:14
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题目大意:
n个点m条无向边的一个图,让你加一条边,使得桥数最少,输出最少桥的个数。
思路:
我们知道,我们希望割边最小的话,我们理所应当先找到各个割边的位子,然后考虑如何在两个连通分量中选两个点,去连这条边。
显然我们可以先将问题转化成一棵树,我们将整个图跑双联通,然后将每个边双联通分量缩成一个点,那么我们就得到了一棵树。那么此时缩点建立了新的图的时候,每条边都是桥。
我们知道,我们想要通过加一条边去使得结果最小的话,我们希望连接两个点之后使得图多出来一个环,那么除了这个环之外的边的个数就是答案。
那么我们显然希望成环部分占用尽可能多的边的个数。那么问题就转化为求树的直径(最长链),那么我们再跑两次Dfs求出树的直径就行了。
代码实现可能稍微有点码农了,但是思路还是很好想的。
Ac代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
using namespace std;
struct node
{
int from,to,next;
}e[150050*3];
int head[150050];
vector<int>mp[150050];
int n,m,sig,cnt,tt,cont,mx,root;
int dfn[150050];
int low[150050];
int vis[150050];
int stack[150050];
int color[150050];
void add(int from,int to)
{
e[cont].to=to;
e[cont].next=head[from];
head[from]=cont++;
}
void Dfs(int u,int from,int depth)
{
vis[u]=1;
if(depth>mx)
{
mx=depth;
root=u;
}
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].to;
if(vis[v]==0)
Dfs(v,u,depth+1);
}
}
void dfs(int u,int from,int depth)
{
vis[u]=1;
mx=max(mx,depth);
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].to;
if(vis[v]==0)
dfs(v,u,depth+1);
}
}
void Work()
{
cont=0;
memset(head,-1,sizeof(head));
for(int i=1;i<=n;i++)
{
for(int j=0;j<mp[i].size();j++)
{
int u=color[i],v=color[mp[i][j]];
if(u==v)continue;
add(u,v);
add(v,u);
}
}
mx=0;
root=1;
memset(vis,0,sizeof(vis));
Dfs(1,-1,0);
mx=0;
memset(vis,0,sizeof(vis));
dfs(root,-1,0);
printf("%d\n",(sig-1)-mx);
}
void Tarjan(int u,int from)
{
vis[u]=1;
dfn[u]=low[u]=cnt++;
stack[++tt]=u;
for(int i=0;i<mp[u].size();i++)
{
int v=mp[u][i];
if(v==from)continue;
if(vis[v]==0)
{
Tarjan(v,u);
low[u]=min(low[u],low[v]);
}
else low[u]=min(low[u],low[v]);
}
if(low[u]==dfn[u])
{
sig++;
do
{
color[stack[tt]]=sig;
vis[stack[tt]]=-1;
}
while(stack[tt--]!=u);
}
}
void Slove()
{
tt=-1;
cnt=1;
sig=0;
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(color,0,sizeof(color));
memset(stack,0,sizeof(stack));
for(int i=1;i<=n;i++)
{
if(vis[i]==0)
{
Tarjan(i,-1);
}
}
Work();
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)mp[i].clear();
for(int i=0;i<m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
mp[x].push_back(y);
mp[y].push_back(x);
}
Slove();
}
}
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