2015 Multi-University Training Contest 2 Friends
Friends
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n−1)2), indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines contains two numbers x and y, which mean x and y are friends. It is guaranteed that x≠y and every friend relationship will appear at most once.
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = ;
struct arc {
int u,v;
} e[maxn];
int st[maxn],du[maxn],n,m,ret;
bool check() {
for(int i = ; i <= n; ++i)
if(st[i]) return false;
return true;
}
bool check2(int x){
if(st[x] == && (du[x]&) == ) return true;
int tmp = st[x]<?du[x]+st[x]:du[x]-st[x];
if(st[x] < && tmp >= && (tmp&) == ) return true;
if(st[x] > && tmp >= && (tmp&) == ) return true;
return false;
}
void dfs(int cur) {
if(cur == m) {
if(check()) ++ret;
return;
}
++st[e[cur].u];
++st[e[cur].v];
--du[e[cur].u];
--du[e[cur].v];
if(check2(e[cur].u) && check2(e[cur].v)) dfs(cur+);
st[e[cur].v] -= ;
st[e[cur].u] -= ;
if(check2(e[cur].u && check2(e[cur].v))) dfs(cur+);
++st[e[cur].v];
++st[e[cur].u];
++du[e[cur].u];
++du[e[cur].v];
}
int main() {
int kase;
scanf("%d",&kase);
while(kase--) {
scanf("%d%d",&n,&m);
memset(du,,sizeof du);
memset(st,,sizeof st);
for(int i = ret = ; i < m; ++i) {
scanf("%d%d",&e[i].u,&e[i].v);
++du[e[i].u];
++du[e[i].v];
}
bool flag = true;
for(int i = ; i <= n && flag; ++i)
if(du[i]&) flag = false;
if(flag) dfs();
printf("%d\n",ret);
}
return ;
}
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