LA3523——无向图的点双连通分量

题意：求一张图中不属于任何一个奇圈的节点的个数。

无向图中，若一个点双连通分量内有一个奇圈，可以看出改连通分量内的所有点都可以包含在一个奇圈内，所以可以用dfs求出图中的所有的点双连通分量，每求出一个连通分量，判断其是否为二分图，若为二分图则该连通分量中没有奇圈，若不是二分图，该连通分量中的所有点都不是要求的点。同时要注意一个点可以同时属于多个点双连通分量。

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <vector>#include <algorithm>using namespace std;const int maxn = 1000 + 10;const int maxm = 1000000 + 10;struct edge { int v, next; }edges[maxm];int n, m, edgehead[maxn], tot, pre[maxn], bccno[maxn], dfs_clock, bcc_cnt, odd[maxn], color[maxn];int now[maxn], top, mar[maxn][maxn];stack<int> S;void init(){    memset(edgehead, 0xff, sizeof(edgehead)); memset(pre, 0, sizeof(pre)); dfs_clock = tot = bcc_cnt = 0;    memset(mar, 0, sizeof(mar));    memset(odd, 0, sizeof(odd));    memset(bccno, 0, sizeof(bccno));}void addedge(int from, int to) { edges[tot].v = to; edges[tot].next = edgehead[from]; edgehead[from] = tot++; }bool bipartite(int u, int b){    for(int i = edgehead[u]; i != -1; i = edges[i].next)    {        int v = edges[i].v; if(bccno[v] != b) continue;        if(color[v] == color[u]) return false;        if(!color[v])        {            color[v] = 3 - color[u];            if(!bipartite(v, b)) return false;        }    }    return true;}int dfs(int u, int fa){    int lowu = pre[u] = ++dfs_clock;    int child = 0;    for(int i = edgehead[u]; i != -1; i = edges[i].next)    {        int v = edges[i].v;        if(!pre[v])        {            S.push(v);            child++;            int lowv = dfs(v, u);            lowu = min(lowu, lowv);            if(lowv >= pre[u])            {                bcc_cnt++; top = 0;                while(1)                {                    int x = S.top(); bccno[x] = bcc_cnt; now[top++] = x;                    if(x != u) S.pop();                    else break;                }                memset(color, 0, sizeof(color));                color[u] = 1;                if(top > 2 && !bipartite(u, bcc_cnt))                    while(top != 0) odd[now[--top]] = 1;            }        }        else if(v != fa) lowu = min(lowu, pre[v]);    }    return lowu;}int main(){    freopen("in", "r", stdin);    while(~scanf("%d %d", &n, &m) && n && m)    {        init();        while(m--)        {            int a, b;            scanf("%d %d", &a, &b);            mar[a][b] = mar[b][a] = 1;        }        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= n; ++j)                if(i != j && !mar[i][j]) addedge(i, j);        for(int i = 1; i <= n; ++i)        {            if(!pre[i])            {                while(!S.empty()) S.pop();                S.push(i); dfs(i, -1);            }        }        int res = 0;        for(int i = 1; i <= n; ++i) if(!odd[i]) res++;        printf("%d\n", res);    }    return 0;}/*5 51 41 52 53 44 50 02*/