2017ACMICPC广西赛C题，如何暴力统计三元环？
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本文共计885个文字，预计阅读时间需要4分钟。
小A是一位天文爱好者，他发现天空如此美丽！于是他开始数星星。天空中共有n颗星星，小A用m条无向边将它们连接起来。保证没有任何边连接同一颗星星。
Little A is an astronomy lover, and he has found that the sky was so beautiful!
So he is counting stars now!
There are n stars in the sky, and little A has connected them by m non-directional edges.
It is guranteed that no edges connect one star with itself, and every two edges connect different pairs of stars.
Now little A wants to know that how many different “A-Structure”s are there in the sky, can you help him?
An “A-structure” can be seen as a non-directional subgraph G, with a set of four nodes V and a set of five edges E.
If V=(A,B,C,D) and E=(AB,BC,CD,DA,AC), we call G as an “A-structure”.
It is defined that “A-structure” G1=V1+E1 and G2=V2+E2 are same only in the condition that V1=V2 and E1=E2.
Input
There are no more than 300 test cases.
For each test case, there are 2 positive integers n and m in the first line.
2≤n≤105, 1≤m≤min(2×105,n(n−1)2)
And then m lines follow, in each line there are two positive integers u and v, describing that this edge connects node u and node v.
1≤u,v≤n
∑n≤3×105,∑m≤6×105
Output
For each test case, just output one integer–the number of different “A-structure”s in one line.
Sample Input
4 5
1 2
2 3
3 4
4 1
1 3
4 6
1 2
2 3
3 4
4 1
1 3
2 4
Sample Output
1
6
暴力找即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 100005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
vector<int>G[maxn];
set<ll>st;
int vis[maxn], lnk[maxn], ot[maxn];
int main() {
ios::sync_with_stdio(false);
int n, m;
ll ans, sum;
while (cin >> n >> m) {
int i, j;
ans = 0; sum = 0;
int part = sqrt(m);
for (i = 1; i <= n; i++)G[i].clear();
ms(vis); st.clear(); ms(lnk); ms(ot);
for (i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
G[u].push_back(v); ot[u]++;
G[v].push_back(u); ot[v]++;
st.insert((ll)u*n + v);
st.insert((ll)v*n + u);
}
for (i = 1; i <= n; i++) {
int x = i;
vis[x] = 1;
for (j = 0; j < G[x].size(); j++) {
lnk[G[x][j]] = x;
}
for (j = 0; j < G[x].size(); j++) {
sum = 0;
int y = G[x][j];
if (vis[y])continue;
if (ot[y] <= part) {
for (int k = 0; k < G[y].size(); k++) {
if (lnk[G[y][k]] == x)sum++;
}
}
else {
for (int k = 0; k < G[x].size(); k++) {
int z = G[x][k];
if (st.find((ll)z*n + y) != st.end())sum++;
}
}
ans += 1ll*(sum - 1)*sum / 2;
}
}
cout << ans << endl;
}
}
本文共计885个文字，预计阅读时间需要4分钟。
小A是一位天文爱好者，他发现天空如此美丽！于是他开始数星星。天空中共有n颗星星，小A用m条无向边将它们连接起来。保证没有任何边连接同一颗星星。
Little A is an astronomy lover, and he has found that the sky was so beautiful!
So he is counting stars now!
There are n stars in the sky, and little A has connected them by m non-directional edges.
It is guranteed that no edges connect one star with itself, and every two edges connect different pairs of stars.
Now little A wants to know that how many different “A-Structure”s are there in the sky, can you help him?
An “A-structure” can be seen as a non-directional subgraph G, with a set of four nodes V and a set of five edges E.
If V=(A,B,C,D) and E=(AB,BC,CD,DA,AC), we call G as an “A-structure”.
It is defined that “A-structure” G1=V1+E1 and G2=V2+E2 are same only in the condition that V1=V2 and E1=E2.
Input
There are no more than 300 test cases.
For each test case, there are 2 positive integers n and m in the first line.
2≤n≤105, 1≤m≤min(2×105,n(n−1)2)
And then m lines follow, in each line there are two positive integers u and v, describing that this edge connects node u and node v.
1≤u,v≤n
∑n≤3×105,∑m≤6×105
Output
For each test case, just output one integer–the number of different “A-structure”s in one line.
Sample Input
4 5
1 2
2 3
3 4
4 1
1 3
4 6
1 2
2 3
3 4
4 1
1 3
2 4
Sample Output
1
6
暴力找即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 100005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
vector<int>G[maxn];
set<ll>st;
int vis[maxn], lnk[maxn], ot[maxn];
int main() {
ios::sync_with_stdio(false);
int n, m;
ll ans, sum;
while (cin >> n >> m) {
int i, j;
ans = 0; sum = 0;
int part = sqrt(m);
for (i = 1; i <= n; i++)G[i].clear();
ms(vis); st.clear(); ms(lnk); ms(ot);
for (i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
G[u].push_back(v); ot[u]++;
G[v].push_back(u); ot[v]++;
st.insert((ll)u*n + v);
st.insert((ll)v*n + u);
}
for (i = 1; i <= n; i++) {
int x = i;
vis[x] = 1;
for (j = 0; j < G[x].size(); j++) {
lnk[G[x][j]] = x;
}
for (j = 0; j < G[x].size(); j++) {
sum = 0;
int y = G[x][j];
if (vis[y])continue;
if (ot[y] <= part) {
for (int k = 0; k < G[y].size(); k++) {
if (lnk[G[y][k]] == x)sum++;
}
}
else {
for (int k = 0; k < G[x].size(); k++) {
int z = G[x][k];
if (st.find((ll)z*n + y) != st.end())sum++;
}
}
ans += 1ll*(sum - 1)*sum / 2;
}
}
cout << ans << endl;
}
}