POJ - 1523 SPF(割点)

来源：互联网发布：dubbo 千兆网络编辑：程序博客网时间：2024/05/01 10:19

题目大意：给出一张图，问去割点后，连通分量的个数有多少

解题思路：割点的水题，套模版就可以
不得不吐槽一下输入。。

#include <cstdio>#include <cstring>#define min(a,b) ((a)<(b)?(a):(b))#define N 1010#define M 2000010struct Edge{    int to, next;}E[M];int head[N], num[N], pre[N], lowlink[N];int tot, dfs_clock, root;bool iscut[N];void dfs(int u, int fa) {    lowlink[u] = pre[u] = ++dfs_clock;    int child = 0;    for (int i = head[u]; i != -1; i = E[i].next) {        int v = E[i].to;        if (!pre[v]) {            child++;            dfs(v, u);            lowlink[u] = min(lowlink[u],lowlink[v]);            if ((u == root && child > 1) || (u != root && pre[u] <= lowlink[v])) {                num[u]++;            }        }        else if (v != fa)  {            lowlink[u] = min(lowlink[u], pre[v]);        }    }}int cas = 1;void solve() {    memset(pre, 0, sizeof(pre));    memset(num, 0, sizeof(num));    dfs_clock = 0;    for (int i = 1; i <= 1000; i++) {        if (!pre[i] && head[i] != -1) {            root = i;            dfs(i,-1);        }    }    printf("Network #%d\n", cas++);    bool flag = false;    for (int i = 1; i <= 1000; i++)        if (num[i]) {            printf("  SPF node %d leaves %d subnets\n", i, num[i] + 1);            flag = true;        }    if (!flag) {        printf("  No SPF nodes\n");    }}void AddEdge(int u, int v) {    E[tot].to = v; E[tot].next = head[u]; head[u] = tot++;    u = u ^ v; v = u ^ v; u = u ^ v;    E[tot].to = v; E[tot].next = head[u]; head[u] = tot++;}int main() {    int a, b;    bool flag = false;    while (scanf("%d", &a) != EOF && a) {        if (flag)            printf("\n");        flag = true;        memset(head, -1, sizeof(head));        tot = 0;        scanf("%d", &b);        AddEdge(a,b);        while (scanf("%d", &a) && a) {            scanf("%d", &b);            AddEdge(a,b);        }        solve();    }    return 0;}

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