标签：input i+1 where highlight ems space nes 描述 line
题目描述
Two undirected simple graphs and where are isomorphic when there exists a bijection on V satisfying if and only if {x, y} ∈ E2.
Given two graphs and , count the number of graphs satisfying the following condition:
* .
* G1 and G are isomorphic.
输入描述:
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains three integers n, m1 and m2 where |E1| = m1 and |E2| = m2.
The i-th of the following m1 lines contains 2 integers ai and bi which denote {ai, bi} ∈ E1.
The i-th of the last m2 lines contains 2 integers ai and bi which denote {ai, bi} ∈ E2.
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入
3 1 2
1 3
1 2
2 3
4 2 3
1 2
1 3
4 1
4 2
4 3
输出
2
3
备注:
* 1 ≤ n ≤ 8
*
* 1 ≤ ai, bi ≤ n
* The number of test cases does not exceed 50.
题意 : 给你两个大小为 N 的图，要求寻找既是B的子图，又和A是同构的图的数量
思路分析：刚开始一直不是很懂它的题意是要干嘛，比赛后也看了很长时间才懂一些，1-2 3 和 1 2-3 是属于同构关系的，因为都是两个点连着，一个点单独出现，因此这里只要N！枚举对应一下就可以了
1 . hash去判重
#include <bits/stdc++.h>
using namespace std;
#define ll unsigned long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
ll n, m1, m2;
bool mp1[10][10], mp2[10][10];
ll arr[10], d[10];
set<ll>s;
void solve() {
for(ll i = 1; i <= n; i++) arr[i] = i;
do{
for(ll i = 1; i <= n; i++) d[i] = arr[i];
ll sum = 0;
for(ll i = 1; i <= n; i++){
for(ll j = i+1; j <= n; j++){
if (mp2[i][j] && mp1[d[i]][d[j]]){
sum++;
}
}
}
if (sum == m1){
ll x = 0;
for(ll i = 1; i <= n; i++){
for(ll j = i+1; j <= n; j++){
if (mp1[d[i]][d[j]]) {
x+=(1LL<<(i*(n-1)+j));
}
}
}
//printf("++++ %llu \n", x);
s.insert(x);
}
}while(next_permutation(arr+1, arr+1+n));
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ll a, b;
while(~scanf("%d%d%d", &n, &m1, &m2)){
memset(mp1, false, sizeof(mp1));
memset(mp2, false, sizeof(mp2));
for(ll i = 1; i <= m1; i++){
scanf("%d%d", &a, &b);
mp1[a][b] = mp1[b][a] = true;
}
for(ll i = 1; i <= m2; i++){
scanf("%d%d", &a, &b);
mp2[a][b] = mp2[b][a] = true;
}
s.clear();
solve();
printf("%d\n", s.size());
}
return 0;
}
2 . 直接这样考虑，就是A图自身的同构在通过B去计算的时候，会重复算的，因此除一下就可以了
using namespace std;
#define ll unsigned long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
ll n, m1, m2;
bool mp1[10][10], mp2[10][10];
ll arr[10], d[10];
set<ll>s;
int a1, a2;
void solve() {
for(ll i = 1; i <= n; i++) arr[i] = i;
do{
for(ll i = 1; i <= n; i++) d[i] = arr[i];
ll sum = 0;
for(ll i = 1; i <= n; i++){
for(ll j = i+1; j <= n; j++){
if (mp2[i][j] && mp1[d[i]][d[j]]){
sum++;
}
}
}
if (sum == m1){
a1++;
}
}while(next_permutation(arr+1, arr+1+n));
}
void solve2() {
for(ll i = 1; i <= n; i++) arr[i] = i;
do{
for(ll i = 1; i <= n; i++) d[i] = arr[i];
ll sum = 0;
for(ll i = 1; i <= n; i++){
for(ll j = i+1; j <= n; j++){
if (mp1[i][j] && mp1[d[i]][d[j]]){
sum++;
}
}
}
if (sum == m1){
a2++;
}
}while(next_permutation(arr+1, arr+1+n));
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ll a, b;
while(~scanf("%d%d%d", &n, &m1, &m2)){
memset(mp1, false, sizeof(mp1));
memset(mp2, false, sizeof(mp2));
for(ll i = 1; i <= m1; i++){
scanf("%d%d", &a, &b);
mp1[a][b] = mp1[b][a] = true;
}
for(ll i = 1; i <= m2; i++){
scanf("%d%d", &a, &b);
mp2[a][b] = mp2[b][a] = true;
}
s.clear();
a1 = a2 = 0;
solve();
solve2();
printf("%d\n", a1/a2);
}
return 0;
}
标签：input i+1 where highlight ems space nes 描述 line
原文地址：https://www.cnblogs.com/ccut-ry/p/9403727.html
