HDU 3926 (图同构)

来源：互联网发布：判断推理知乎编辑：程序博客网时间：2024/04/30 02:39
题目链接:点击这里

题意:给出两个最大度数是2的无向图, 判断是否同构.

因为最大度数是2, 直接把所有的环和链抓出来分别判断相等就好了.

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <vector>#include <cstring>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define maxn 111111#define maxm 411111struct E {    int v, next;}edge1[maxm], edge2[maxm];int head1[maxn], head2[maxn];vector <int> ans1[2], ans2[2];int cnt1, cnt2;int n1, n2, m1, m2;bool vis[maxn], is_loop;int len;void add_edge (int u, int v, int op) {    if (op == 1) {        edge1[cnt1].v = v, edge1[cnt1].next = head1[u], head1[u] = cnt1++;    }    else {        edge2[cnt2].v = v, edge2[cnt2].next = head2[u], head2[u] = cnt2++;    }}void dfs1 (int u, int fa) {     vis[u] = 1;    len++;    for (int i = head1[u]; i != -1; i = edge1[i].next) {        int v = edge1[i].v;        if (v == fa) continue;        if (vis[v]) {            is_loop = 1;            continue;        }        dfs1 (v, u);    }}void dfs2 (int u, int fa) {    vis[u] = 1;    len++;    for (int i = head2[u]; i != -1; i = edge2[i].next) {        int v = edge2[i].v;        if (v == fa) continue;        if (vis[v]) {            is_loop = 1;            continue;        }        dfs2 (v, u);    }}bool ok () {    sort (ans1[0].begin (), ans1[0].end ()); sort (ans1[1].begin (), ans1[1].end ());    sort (ans2[0].begin (), ans2[0].end ()); sort (ans2[1].begin (), ans2[1].end ());    if (ans1[0].size () != ans2[0].size () || ans1[1].size () != ans2[1].size ())        return 0;    int Max = ans1[0].size ();    for (int i = 0; i < Max; i++) if (ans1[0][i] != ans2[0][i]) return 0;    Max = ans1[1].size ();    for (int i = 0; i < Max; i++) if (ans1[1][i] != ans2[1][i]) return 0;    return 1;}int main () {    int t, kase = 0;    scanf ("%d", &t);    while (t--) {        scanf ("%d%d", &n1, &m1);        memset (head1, -1, sizeof head1);        memset (head2, -1, sizeof head2);        cnt1 = 0;        for (int i = 0; i < m1; i++) {            int u, v;            scanf ("%d%d", &u, &v);            add_edge (u, v, 1);            add_edge (v, u, 1);        }        scanf ("%d%d", &n2, &m2);        cnt2 = 0;        for (int i = 0; i < m2; i++) {            int u, v;            scanf ("%d%d", &u, &v);            add_edge (u, v, 2);            add_edge (v, u, 2);        }        printf ("Case #%d: ", ++kase);        if (n1 != n2 || m1 != m2) {            printf ("NO\n");            continue;        }        ans1[0].clear (), ans1[1].clear (),  ans2[0].clear (), ans2[1].clear ();        memset (vis, 0, sizeof vis);        for (int i = 1; i <= n1; i++) if (!vis[i]) {            len = 0;            is_loop = 0;            dfs1 (i, 0);             if (is_loop) {                ans1[1].push_back (len);            }            else                ans1[0].push_back (len);        }        memset (vis, 0, sizeof vis);        for (int i = 1; i <= n1; i++) if (!vis[i]) {            len = 0;            is_loop = 0;            dfs2 (i, 0);            if (is_loop) {                ans2[1].push_back (len);            }            else                ans2[0].push_back (len);        }        printf ("%s\n", ok () ? "YES" : "NO");    }    return 0;}
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