hdu3844 Mining Your Own Business，無向圖的雙連通分量

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無向圖的雙連通分量

#include<cstdio>#include<stack>#include<vector>#include<map>#include<algorithm>#include<cstring>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;const int maxn = 50010;struct Edge {    int u, v;    Edge(int u, int v):u(u), v(v) {}};int pre[maxn], low[maxn], bccno[maxn], iscut[maxn], bcc_cnt, dfs_clock;vector<int> g[maxn], bcc[maxn];stack<Edge> S;int dfs(int u, int fa){    int lowu = pre[u] = ++dfs_clock;    int child = 0;    for(int i = 0; i < g[u].size(); i++) {        int v = g[u][i];        Edge e = Edge(u, v);        if(!pre[v]) {            S.push(e);            child++;            int lowv = dfs(v, u);            lowu = min(lowu, lowv);            if(lowv >= pre[u]) {                iscut[u] = 1;                bcc_cnt++;                bcc[bcc_cnt].clear();                for(;;) {                    Edge x = S.top();                    S.pop();                    if(bccno[x.u] != bcc_cnt) {                        bccno[x.u] = bcc_cnt;                        bcc[bcc_cnt].push_back(x.u);                    }                    if(bccno[x.v] != bcc_cnt) {                        bccno[x.v] = bcc_cnt;                        bcc[bcc_cnt].push_back(x.v);                    }                    if(x.u == u && x.v == v) break;                }            }        } else if(pre[v] < pre[u] && v!= fa) {            S.push(e);            lowu = min(lowu, pre[v]);        }    }    if(child == 1 && fa < 0) iscut[u] = 0;    return low[u] = lowu;}void find_bcc(int n){    memset(iscut, 0, sizeof(iscut));    memset(pre, 0, sizeof(pre));    memset(bccno, 0, sizeof(bccno));    dfs_clock = bcc_cnt = 0;    for(int i = 0; i < n; i++)        if(!pre[i]) dfs(i, -1);}int kase;void solve(int n){    find_bcc(n);    LL ans1 = 0, ans2 = 1;    for(int i = 1; i <= bcc_cnt; i++) {        int cut_cnt = 0;        for(int j = 0; j < bcc[i].size(); j++)            if(iscut[bcc[i][j]]) cut_cnt++;        if(cut_cnt == 1)            ans1++, ans2 *= (LL)(bcc[i].size() - cut_cnt);    }    if(bcc_cnt == 1) {        ans1 = 2, ans2 = (LL)(n-1)*n/2;    }    printf("Case %d: %I64d %I64d\n", kase, ans1, ans2);}int main(void){    int m;    while(scanf("%d", &m), m) {        kase++;        for(int i = 0; i < maxn; i++)            g[i].clear();        int mxn = 0;        while(m--) {            int u, v;            scanf("%d%d", &u, &v);            mxn = max(mxn, max(u, v));            u--, v--;            g[u].push_back(v), g[v].push_back(u);        }        solve(mxn);    }    return 0;}

hdu3844 Mining Your Own Business，無向圖的雙連通分量