POJ 3352 Road Construction （縮點）__圖論

Description

It’s almost summer time, and that means that it’s almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

  Input

The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

  Output

One line, consisting of an integer, which gives the minimum number of roads that we need to add.

  Sample Input

10 121 21 31 42 52 65 63 73 87 84 94 109 10

  Sample Output

2

  題意

給出一張圖，問至少增加多少條邊才能使其成為強連通圖。

  思路

首先我們可以利用 Tarjan 演算法求出圖中所有的強連通分量，然後根據每個點所在的強連通分量進行縮點。

縮點的結果是一棵樹，於是問題便轉化為了求在該縮點樹上至少需要增加多少條邊才能使其強連通。

顯然其結果是：(樹中度為 1 的節點數目 + 1) / 2

  AC 代碼

#include<iostream>#include<cstdlib>#include<stdio.h>#include<cstring>using namespace std;const int maxn = 2e3;struct node{    int to;    int next;} e[maxn];int head[maxn],tot;int n,m;int dfn[maxn],low[maxn],ind,ans;bool instack[maxn];int Stack[maxn],Stop,Bcnt;int blong[maxn];bool vis[maxn][maxn];bool viss[maxn][maxn];void init(){    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(vis,false,sizeof(vis));    memset(instack,false,sizeof(instack));    tot=ind=ans=Stop=Bcnt=0;}void addedge(int u,int v){    e[tot].to=v;    e[tot].next=head[u];    head[u]=tot++;}void dfs(int x,int fa){    dfn[x]=low[x]=++ind;    instack[x]=true;    Stack[++Stop]=x;    for(int i=head[x]; i!=-1; i=e[i].next)    {        int to=e[i].to;        if(to==fa)continue;        if(!dfn[to])        {            dfs(to,x);            if(low[to]<low[x])                low[x]=low[to];        }        else if(instack[to]&&dfn[to]<low[x])            low[x]=dfn[to];    }    if (dfn[x]==low[x]) //相等說明i為強連通子圖搜尋樹的根節點    {        Bcnt++;        int top;        do  //退棧        {            top=Stack[Stop--];            blong[top]=Bcnt;        //記錄每一個點屬於哪一個強連通分量            instack[top]=false;        }        while (top!=x);    }}void dfs2(int x,int fa)     //統計度為 1 地節點個數{    int num=0;    dfn[x]=1;    for(int i=head[x]; i!=-1; i=e[i].next)    {        num++;        int to=e[i].to;        if(to==x||dfn[to])continue;        dfs2(to,x);    }    if(num==1)ans++;}int main(){    while(~scanf("%d%d",&n,&m))    {        init();        for(int i=0; i<m; i++)        {            int u,v;            scanf("%d%d",&u,&v);            viss[u][v]=viss[v][u]=true;            addedge(u,v);            addedge(v,u);        }        memset(blong,0,sizeof(blong));        dfs(1,-1);        init();        for(int i=1; i<=n; i++)            for(int j=i+1; j<=n; j++)   //如果 i,j 有邊，並且它們屬於兩個不同的強連通分量，說明 i<->j 為橋            {                if(viss[i][j]&&blong[i]!=blong[j]&&!vis[blong[i]][blong[j]])                {                    vis[blong[i]][blong[j]]=true;                    addedge(blong[i],blong[j]);     //縮點重建立圖                    addedge(blong[j],blong[i]);                }            }        dfs2(1,-1);        printf("%d\n",(ans+1)/2);    }    return 0;}