poj 2117 Electricity（tarjan求割点删掉之后的连通块数）

Posted 2020-09-19 Gealo

题目链接：http://poj.org/problem?id=2117

题意：求删除一个点后，图中最多有多少个连通块。

 

题解：就是找一下割点，根节点的割点删掉后增加son-1（son为子树个数），非根节点删掉之后++

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1e4 + 10;
const int M = 1e6 + 10;
struct TnT {
    int v , next;
    bool cut;
}edge[M];
int head[N] , e;
int Low[N] , DFN[N] , Stack[N] , add_block[N];
bool Instack[N];
bool cut[N];
int Index , bridge , top;
void init() {
    memset(head , -1 , sizeof(head));
    e = 0;
}
void add(int u , int v) {
    edge[e].v = v , edge[e].next = head[u] ,edge[e].cut = false , head[u] = e++;
}
void Tarjan(int u , int pre) {
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    int son = 0;
    for(int i = head[u] ; i != -1 ; i = edge[i].next) {
        v = edge[i].v;
        if(v == pre) continue;
        if(!DFN[v]) {
            son++;
            Tarjan(v , u);
            Low[u] = min(Low[u] , Low[v]);
            if(Low[v] > DFN[u]) {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
            if(u != pre && Low[v] >= DFN[u]) {
                cut[u] = true;
                add_block[u]++;
            }
        }
        else if(Instack[v]) Low[u] = min(Low[u] , DFN[v]);
    }
    if(u == pre && son > 1) cut[u] = true;
    if(u == pre) add_block[u] = son - 1;
    Instack[u] = false;
    top--;
}
int main() {
    int p , c;
    while(~scanf("%d%d" , &p , &c)) {
        if(p == 0 && c == 0) break;
        init();
        for(int i = 0 ; i < c ; i++) {
            int u , v;
            scanf("%d%d" , &u , &v);
            add(u , v);
            add(v , u);
        }
        memset(DFN , 0 , sizeof(DFN));
        memset(Instack , false , sizeof(Instack));
        memset(add_block , 0 , sizeof(add_block));
        memset(cut , false , sizeof(cut));
        int cnt = 0;
        Index = 0 , bridge = 0 , top = 0;
        for(int i = 0 ; i < p ; i++) {
            if(!DFN[i]) {
                Tarjan(i , i) , cnt++;
            }
        }
        int MAX = 0;
        for(int i = 0 ; i < p ; i++) MAX = max(MAX , cnt + add_block[i]);
        printf("%d\n" , MAX);
    }
    return 0;
}