POJ 2117--Electricity【點雙聯通 && 求刪去一個點後，圖最多有多少塊連通】

標籤：

Electricity

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4589		Accepted: 1512

Description

Blackouts and Dark Nights (also known as ACM++) is a company that provides electricity. The company owns several power plants, each of them supplying a small area that surrounds it. This organization brings a lot of problems - it often happens that there is not enough power in one area, while there is a large surplus in the rest of the country. 

ACM++ has therefore decided to connect the networks of some of the plants together. At least in the first stage, there is no need to connect all plants to a single network, but on the other hand it may pay up to create redundant connections on critical places - i.e. the network may contain cycles. Various plans for the connections were proposed, and the complicated phase of evaluation of them has begun. 

One of the criteria that has to be taken into account is the reliability of the created network. To evaluate it, we assume that the worst event that can happen is a malfunction in one of the joining points at the power plants, which might cause the network to split into several parts. While each of these parts could still work, each of them would have to cope with the problems, so it is essential to minimize the number of parts into which the network will split due to removal of one of the joining points. 

Your task is to write a software that would help evaluating this risk. Your program is given a description of the network, and it should determine the maximum number of non-connected parts from that the network may consist after removal of one of the joining points (not counting the removed joining point itself). 

Input

The input consists of several instances. 

The first line of each instance contains two integers 1 <= P <= 10 000 and C >= 0 separated by a single space. P is the number of power plants. The power plants have assigned integers between 0 and P - 1. C is the number of connections. The following C lines of the instance describe the connections. Each of the lines contains two integers 0 <= p1, p2 < P separated by a single space, meaning that plants with numbers p1 and p2 are connected. Each connection is described exactly once and there is at most one connection between every two plants. 

The instances follow each other immediately, without any separator. The input is terminated by a line containing two zeros. 

Output

The output consists of several lines. The i-th line of the output corresponds to the i-th input instance. Each line of the output consists of a single integer C. C is the maximum number of the connected parts of the network that can be obtained by removing one of the joining points at power plants in the instance.

Sample Input

3 30 10 22 14 20 12 33 11 00 0

Sample Output

122

求刪除一個點後， 圖中最多有多少個連通塊

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#define maxn 100000+10#define maxm 2000000+10using namespace std;int n ,m;struct node {    int u, v, next;};node edge[maxm];int head[maxn], cnt;int low[maxn];//從該點或它的子孫出發 通過回邊可以到達的最低int dfn[maxn];//該點的深度優先數bool is_cut[maxn];//標記該點是不是割點int add_bcc[maxn];//去掉該點增加的bcc數目int dfs_clock;//深度優先數計數器int bccno[maxn];//屬於哪個bccint bcc_cnt;//bcc計數器int num;void init(){    cnt = 0;    memset(head, -1, sizeof(head));}void add(int u, int v){    edge[cnt] = {u, v, head[u]};    head[u] = cnt++;    edge[cnt] = {v, u, head[v]};    head[v] = cnt++;}void getmap(){    while(m--){        int a, b;        scanf("%d%d", &a, &b);        a++, b++;        add(a, b);    }}void tarjan(int u, int fa){    low[u] = dfn[u] = ++dfs_clock;    int son = 0;    for(int i = head[u]; i != -1; i = edge[i].next){        int v = edge[i].v;        if(!dfn[v]){            son++;            tarjan(v, u);            low[u] = min(low[u], low[v]);            if( u != fa && low[v] >= dfn[u]){ // u 是割點，但不是根節點                is_cut[u] = true;                //記錄 u 有幾個滿足low[v] >= dfn[u]的子節點 v,                //若有d個，去掉該割點後分成了 d + 1個連通分量,新增了d個                add_bcc[u]++;                //記錄bcc裡面的點            }        }        else if(v != fa && dfn[v] < dfn[u]){            low[u] = min(low[u], dfn[v]);//回邊更新        }    }    //u 是根節點， 但它的子節點只有一個，不是割點    if(u == fa && son == 1) is_cut[u] = 0;    //u 是根節點， 而且它的子節點個數 >= 2,是割點。刪去該點後    //u 有幾個子節點， 就分成了幾個bcc，新增了son - 1    if(u == fa && son > 1) add_bcc[u] = son - 1;}void find(){    memset(low, 0, sizeof(low));    memset(dfn, 0, sizeof(dfn));    memset(add_bcc, 0, sizeof(add_bcc));    memset(is_cut, 0, sizeof(is_cut));    memset(bccno, 0, sizeof(bccno));    dfs_clock = bcc_cnt = 0;    num = 0;//記錄原來的聯通塊    for(int i = 1; i <= n; ++i)        if(!dfn[i]){            tarjan(i, i);            num++;        }}void solve(){    num = 0;    find();    int ans = 0;    //printf("***%d\n", num);    for(int i = 1; i <= n; ++i){    //printf("---%d\n", add_bcc[i]);        ans = max(ans, add_bcc[i]);    }    printf("%d\n", ans + num);}int main (){    while(scanf("%d%d", &n, &m), n || m){    if(m == 0){    printf("%d\n", n - 1);    continue;}        init();        getmap();        solve();    }    return 0;}

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POJ 2117--Electricity【點雙聯通 && 求刪去一個點後，圖最多有多少塊連通】