LightOJ - 1063 Ant Hills（割点）

来源：互联网发布：unity 引擎源码编辑：程序博客网时间：2024/05/23 20:55
题目大意：求无向图中，有多少个割点
解题思路：模版题了
#include <cstdio>#include <cstring>#include <vector>#include <stack>using namespace std;#define max(a,b)((a)>(b)?(a):(b))#define min(a,b)((a)<(b)?(a):(b))const int MAXNODE = 10005;const int MAXEDGE = 100010;const int INF = 0x3f3f3f3f;struct Edge{    int v, id, next;    bool bridge;    Edge() {}    Edge(int v, int id, int next): v(v), id(id), next(next), bridge(false){}}E[MAXEDGE * 2], cut[MAXEDGE];struct Node {    int u, v;    Node() {}    Node(int u, int v): u(u), v(v) {}};//bcc表示的是一个bcc里面的点vector<int> bcc[MAXNODE];stack<Node> Stack;//pre纪录的是时间戳,lowlink纪录的是该点及其该子孙节点所能返回的最早时间戳是多少,bccno纪录的是该点当前是属于哪个bcc的int head[MAXNODE], pre[MAXNODE], lowlink[MAXNODE], bccno[MAXNODE];int n, m, tot, bcc_cnt, dfs_clock, cut_cnt; bool iscut[MAXNODE];void AddEdge(int u, int v, int id) {    E[tot] = Edge(v, id, head[u]);    head[u] = tot++;    u = u ^ v; v = u ^ v; u = u ^ v;    E[tot] = Edge(v, id, head[u]);    head[u] = tot++;}void init() {    scanf("%d%d", &n, &m);    memset(head, -1, sizeof(head));    tot = 0;    int u, v;    for (int i = 0; i < m; i++) {        scanf("%d%d", &u, &v);        AddEdge(u, v, 0);    }}//点双连通void dfs(int u, int fa) {    pre[u] = lowlink[u] = ++dfs_clock;    int child = 0;//纪录当前节点有多少个子节点    for (int i = head[u]; ~i; i = E[i].next) {        int v = E[i].v;        if (!pre[v]) {            Stack.push(Node(u, v));            child++;            dfs(v, u);            lowlink[u] = min(lowlink[u], lowlink[v]);//更新            //子节点最多返回到该点            if (lowlink[v] >= pre[u]) {                //该边为桥                if (lowlink[v] > pre[u]) {                    E[i].bridge = E[i ^ 1].bridge = true;                    cut[cut_cnt++] = E[i];                }                iscut[u] = true;                bcc_cnt++; bcc[bcc_cnt].clear();                while (1) {                    Node x = Stack.top(); Stack.pop();                    if (bccno[x.u] != bcc_cnt) {                        bcc[bcc_cnt].push_back(x.u);                        bccno[x.u] = bcc_cnt;                    }                    if (bccno[x.v] != bcc_cnt) {                        bcc[bcc_cnt].push_back(x.v);                        bccno[x.v] = bcc_cnt;                    }                    if (x.u == u && x.v == v) break;                }            }        }        else if (v != fa && pre[v] < pre[u]) {//反向边            Stack.push(Node(u, v));            lowlink[u] = min(lowlink[u], pre[v]);        }    }    //u是根结点，且只有一个孩子，那就不是割点了    if (fa < 0 && child == 1) iscut[u] = 0;}/*//边双连通int belong[MAXNODE];int Stack[MAXNODE];void find_bcc(int u, int fa) {    pre[u] = lowlink[u] = ++dfs_clock;    Stack[++top] = u;    for (int i = head[u]; ~i; i = E[i].next) {        int v = E[i].v;        if (!pre[v]) {            find_bcc(v, u);            lowlink[u] = min(lowlink[u], lowlink[v]);            if (lowlink[v] > pre[u]) {                 E[i].bridge = E[i ^ 1].bridge = true;                cut[cut_cnt++] = E[i];                bcc_cnt++;                while (1) {                    int x = Stack[top--];                    belong[x] = bcc_cnt;                    if (x == v) break;                }            }        }        else if (pre[v] < pre[u] && v != fa) lowlink[u] = min(lowlink[u], pre[v]);    }}*/void find_bcc() {    memset(pre, 0, sizeof(pre));    memset(iscut, 0, sizeof(iscut));    memset(bccno, 0, sizeof(bccno));    dfs_clock = bcc_cnt = cut_cnt = 0;    for (int i = 1; i <= n; i++)         if (!pre[i]) dfs(i, -1);}int cas = 1;void solve() {    find_bcc();    int ans = 0;    for (int i = 1; i <= n; i++)        if (iscut[i]) ans++;    printf("Case %d: %d\n", cas++, ans);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}
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