(POJ 3694) Network 求橋個數

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題目連結:http://poj.org/problem?id=3694

DescriptionA network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can‘t be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.You are to help the administrator by reporting the number of bridges in the network after each new link is added.InputThe input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.The last test case is followed by a line containing two zeros.OutputFor each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.Sample Input3 21 22 321 21 34 41 22 12 31 421 23 40 0Sample OutputCase 1:10Case 2:20

題目大意:有N個點,M條邊,添加Q條邊,問每填一條邊,還有幾條橋?

#include<stdio.h>#include<cstdlib>#include<cstring>#include<algorithm>#include<queue>#include<math.h>#include <stack>using namespace std;#define ll long long#define INF 0x3f3f3f3f#define met(a,b) memset(a,b,sizeof(a))#define mod 2147493647#define N 100100int low[N],dfn[N],vis[N];int fa[N];int n,t,ans;int bin[N];vector<vector<int> >Q;void init(){    met(low,0);    met(dfn,0);    t=0;    Q.clear();    Q.resize(n+10);    met(bin,0);    met(fa,0);}void tarjin(int u,int f){    low[u]=dfn[u]=++t;    fa[u]=f;    int l=Q[u].size();    for(int i=0; i<l; i++)    {        int v=Q[u][i];        if(!dfn[v])        {            tarjin(v,u);            low[u]=min(low[u],low[v]);            if(dfn[u]<low[v])///開始沒被尋找,可能是橋            {                bin[v]++;                ans++;            }        }        else if(f!=v)        {            low[u]=min(low[u],dfn[v]);            if(dfn[u]<low[v])///被尋找過,有兩條路通這個點,不是橋            {                bin[v]--;                ans--;            }        }    }}void LCR(int a,int b){    if(a==b)        return ;    if(dfn[a]<dfn[b])    {        if(bin[b]==1)        {            bin[b]=0;            ans--;        }        LCR(a,fa[b]);    }    else    {        if(bin[a]==1)        {            bin[a]=0;            ans--;        }        LCR(fa[a],b);    }}int main(){    int m,e,f;    int cot=1,q;    while(scanf("%d %d",&n,&m),n+m)    {        init();        ans=0;        for(int i=0; i<m; i++)        {            scanf("%d %d",&e,&f);            Q[e].push_back(f);            Q[f].push_back(e);        }        tarjin(1,-1);        scanf("%d",&q);        printf("Case %d:\n",cot++);        while(q--)        {            scanf("%d %d",&e,&f);            LCR(e,f);            printf("%d\n",ans);        }    }    return 0;}

 

(POJ 3694) Network 求橋個數

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