Hdu-4612-Strongly Connected Component

Question:

There are some small islands with some edges that allow you to add one edge to minimize the number of bridges on those sides.

Practice:
1. Map the island as a vertex and the island connected to the edge.

2. Find all the strongly connected components in the graph.

3. Regard All the strong Unicom components as a point.

4. Find the diameter of the tree. The newly added side should be on both sides of the diameter to minimize the bridge in the graph.

 

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<stdio.h>#include<algorithm>#include<vector>#include<queue>#include<stack>#include<string.h>using namespace std;#define maxn 200003#define mem(a,b) memset(a,b,sizeof(a))vector<int>q[maxn];stack<int>qq;struct list{    int e;    int next;}edge[maxn*10];struct ll{    int u;    int v;}node[maxn*5];int tops,times,nums,n,m;int dnf[maxn];int low[maxn];int instack[maxn];int head[maxn*10];int vis[maxn*10];int dist[maxn];int num[maxn];int visit[maxn];void add(int x,int y){    edge[tops].e=y;    edge[tops].next=head[x];    head[x]=tops++;}void tarjan(int x){    dnf[x]=low[x]=times++;    instack[x]=1;    qq.push(x);    for(int i=head[x];i!=-1;i=edge[i].next)    {        int y=edge[i].e;        if(vis[i])continue;        vis[i]=vis[i^1]=1;        if(!dnf[y])        {            tarjan(y);            low[x]=min(low[x],low[y]);        }        else if(instack[y])        {            low[x]=min(low[x],dnf[y]);        }    }    if(low[x]==dnf[x])    {        int y=-1;        while(x!=y)        {            y=qq.top();            qq.pop();            instack[y]=0;            num[y]=nums;        }        nums++;    }}int spfa(int x){    for(int i=0;i<=nums;i++)    {        dist[i]=-1;        visit[i]=0;    }    queue<int>pp;    pp.push(x);    dist[x]=0;    visit[x]=1;    while(!pp.empty())    {        int y=pp.front();        pp.pop();        for(int i=0;i<q[y].size();i++)        {            int z=q[y][i];            if(!visit[z])            {                dist[z]=dist[y]+1;                visit[z]=1;                pp.push(z);            }        }    }    int maxx,ip;    maxx=-1;    for(int i=0;i<nums;i++)    {        if(dist[i]>maxx)        {            maxx=dist[i];            ip=i;        }    }    return ip;}int main(){    int a,b,i;    while(scanf("%d%d",&n,&m)&&(n||m))    {        tops=0;        times=1;        nums=0;        for(i=0;i<=m*2;i++)        {            vis[i]=0;            head[i]=-1;        }        for(i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            add(a,b);            add(b,a);            node[i].u=a;            node[i].v=b;        }        for(i=0;i<=n;i++)dnf[i]=low[i]=instack[i]=0;        while(!qq.empty())qq.pop();        for(i=1;i<=n;i++)            if(!dnf[i])tarjan(i);        tops=0;        for(i=0;i<=nums;i++)q[i].clear();        for(i=0;i<m;i++)        {            if(num[node[i].u]!=num[node[i].v])            {                q[num[node[i].u]].push_back(num[node[i].v]);                q[num[node[i].v]].push_back(num[node[i].u]);            }        }        printf("%d\n",nums-dist[spfa(spfa(0))]-1);    }    return 0;}