hdu 4612 Warm up
Warm up
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2012 Accepted Submission(s): 474
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.
思路：题目要求的是加一条边使得剩下的桥的数量最少，我们可以先求原图的双连通分量，缩点后得到一棵树，树的每条边都是原图中的桥，那么我们可以求这个树的最长链，将最长链的两端加一条边后所减少的桥是最多的，所以剩下的桥显然是最少的，所以原题目就是双连通分量缩点并且记录桥的个数，缩点后求树的最长链，答案则是桥的数量减去最长链的长度。看了下大部分过这道题的都是交的C++，貌似G++不能手动扩栈，导致交G++的大部分都爆栈了。下面是代码实现，仅供参考。。。
#pragma comment(linker, "/STACK:102400000,102400000")
#include <string.h>
#include <stdio.h>
#include <algorithm>
#define maxn 200010
struct Edge
{
int to;
int num;
int next;
}e[2][2000100];
int box[2][maxn],cnt[2];
void init()
{
memset(box,-1,sizeof(box));
cnt[0]=cnt[1]=0;
}
void add(int from,int to,int num,int tt)
{
e[tt][cnt[tt]].to=to;
e[tt][cnt[tt]].num=num;
e[tt][cnt[tt]].next=box[tt][from];
box[tt][from]=cnt[tt]++;
}
int pre[maxn];
int low[maxn];
int bridge[1000010];
int bcnt=0;
int cnt0;
void bridge_search(int now,int edge)
{
int t;
int v,w;
low[now]=pre[now]=++cnt0;
for(t=box[0][now];t+1;t=e[0][t].next)
{
v=e[0][t].to;
if(edge!=-1&&e[0][t].num==e[0][edge].num)
continue;
if(!pre[v])
{
bridge_search(v,t);
if(low[v]<low[now])
low[now]=low[v];
if(low[v]>pre[now])
{
if(bridge[e[0][t].num]==0)
bcnt++;
bridge[e[0][t].num]=1;
}
}
else
{
if(low[now]>pre[v])
low[now]=pre[v];
}
}
}
int Bridge(int n)
{
int i;
cnt0=0;
memset(pre,0,sizeof(pre));
memset(low,0,sizeof(low));
memset(bridge,0,sizeof(bridge));
bcnt=0;
for(i=1;i<=n;i++)
{
if(!pre[i])
{
bridge_search(i,-1);
}
}
return bcnt;
}
int vis[maxn];
int dist[maxn];
void Dfs(int now,int num)
{
low[now]=num;
vis[now]=1;
int t,v,nn;
for(t=box[0][now];t+1;t=e[0][t].next)
{
v=e[0][t].to,nn=e[0][t].num;
if(bridge[nn])
continue;
if(!vis[v])
{
Dfs(v,num);
}
}
}
void lensolve(int now)
{
int t,v;
for(t=box[1][now];t+1;t=e[1][t].next)
{
v=e[1][t].to;
if(dist[v]==-1)
{
dist[v]=dist[now]+1;
lensolve(v);
}
}
}
void solve(int n,int num)
{
int i,sum=0;
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
{
if(!low[i])
Dfs(i,++sum);
}
for(i=1;i<=n;i++)
{
int t,v;
for(t=box[0][i];t+1;t=e[0][t].next)
{
v=e[0][t].to;
if(low[i]!=low[v])
{
add(low[i],low[v],0,1);
add(low[v],low[i],0,1);
}
}
}
if(sum==1)
{
printf("0\n");
return;
}
memset(dist,-1,sizeof(dist));
dist[1]=0;
lensolve(1);
int ma=0,root;
for(i=1;i<=sum;i++)
{
if(ma<dist[i])
{
ma=dist[i];
root=i;
}
}
memset(dist,-1,sizeof(dist));
dist[root]=0;
lensolve(root);
ma=0;
for(i=1;i<=sum;i++)
{
if(ma<dist[i])
{
ma=dist[i];
}
}
printf("%d\n",num-ma);
}
int main()
{
// freopen("dd.txt","r",stdin);
int n,m;
while(scanf("%d%d",&n,&m)&&(n+m))
{
init();
int x,y,i;
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
if(x!=y)
{
add(x,y,i,0);
add(y,x,i,0);
}
}
int sum=Bridge(n);
solve(n,sum);
}
return 0;
}
#pragma comment(linker, "/STACK:102400000,102400000")
#include <string.h>
#include <stdio.h>
#include <algorithm>
#define maxn 200010
struct Edge
{
int to;
int num;
int next;
}e[2][2000100];
int box[2][maxn],cnt[2];
void init()
{
memset(box,-1,sizeof(box));
cnt[0]=cnt[1]=0;
}
void add(int from,int to,int num,int tt)
{
e[tt][cnt[tt]].to=to;
e[tt][cnt[tt]].num=num;
e[tt][cnt[tt]].next=box[tt][from];
box[tt][from]=cnt[tt]++;
}
int pre[maxn];
int low[maxn];
int bridge[1000010];
int bcnt=0;
int cnt0;
void bridge_search(int now,int edge)
{
int t;
int v,w;
low[now]=pre[now]=++cnt0;
for(t=box[0][now];t+1;t=e[0][t].next)
{
v=e[0][t].to;
if(edge!=-1&&e[0][t].num==e[0][edge].num)
continue;
if(!pre[v])
{
bridge_search(v,t);
if(low[v]<low[now])
low[now]=low[v];
if(low[v]>pre[now])
{
if(bridge[e[0][t].num]==0)
bcnt++;
bridge[e[0][t].num]=1;
}
}
else
{
if(low[now]>pre[v])
low[now]=pre[v];
}
}
}
int Bridge(int n)
{
int i;
cnt0=0;
memset(pre,0,sizeof(pre));
memset(low,0,sizeof(low));
memset(bridge,0,sizeof(bridge));
bcnt=0;
for(i=1;i<=n;i++)
{
if(!pre[i])
{
bridge_search(i,-1);
}
}
return bcnt;
}
int vis[maxn];
int dist[maxn];
void Dfs(int now,int num)
{
low[now]=num;
vis[now]=1;
int t,v,nn;
for(t=box[0][now];t+1;t=e[0][t].next)
{
v=e[0][t].to,nn=e[0][t].num;
if(bridge[nn])
continue;
if(!vis[v])
{
Dfs(v,num);
}
}
}
void lensolve(int now)
{
int t,v;
for(t=box[1][now];t+1;t=e[1][t].next)
{
v=e[1][t].to;
if(dist[v]==-1)
{
dist[v]=dist[now]+1;
lensolve(v);
}
}
}
void solve(int n,int num)
{
int i,sum=0;
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
{
if(!low[i])
Dfs(i,++sum);
}补充：软件开发 , C++ ,
