标签：contain cto scanf 一个 += 个数 不能 algo ati
Given a simple graph, output the number of simple cycles in it. A simple cycle is a cycle with no repeated vertices or edges.
Input
The first line of input contains two integers n and m (1 ≤ n ≤ 19, 0 ≤ m) – respectively the number of vertices and edges of the graph. Each of the subsequent mlines contains two integers a and b, (1 ≤ a, b ≤ n, a ≠ b) indicating that vertices aand b are connected by an undirected edge. There is no more than one edge connecting any pair of vertices.
Output
Output the number of cycles in the given graph.
Example
4 6
1 2
1 3
1 4
2 3
2 4
3 4
7
Note
The example graph is a clique and contains four cycles of length 3 and three cycles of length 4.
1-2-3 2-3-4 1-3-4 1-2-4 1-2-3-4 1-2-4-3 1-4-2-3
题意:给出一个图的点数和边数输出这个图中有几个环.
题解:这道题可以很容易想到状压,因为数据也只有19(orz).用sta的二进制表示已经有几个点在这个状态中,那怎么表示形成环呢?只需要找到一个当前状态中的点,它神奇的与前面的某一个点有一条边连着,那么说明肯定能构成环,可以为总答案做贡献.如果不能,那么就为下一个状态提供个数.不过由于是无向图.所以两个点也会被认为成环(两个点构成环的个数为边数),并且每个更大的环都会被计算两次,所以最后要减掉.然后就搞定了.
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; long long dp[1<<20][20],ans=0; vector<int> e[20]; int lowbit(int x) { return x&(-x); } int main() { int n,m,f,t; scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { scanf("%d%d",&f,&t); e[f-1].push_back(t-1); e[t-1].push_back(f-1); } for(int i=0;i<n;i++) { dp[1<<i][i]=1; } for(int sta=1;sta<(1<<n);sta++) { for(int i=0;i<n;i++) { if(dp[sta][i]) { for(int k=0;k<e[i].size();k++) { int j=e[i][k]; if(lowbit(sta)>(1<<j)) { continue; } if(sta&(1<<j)) { if(lowbit(sta)==(1<<j)) { ans+=dp[sta][i]; } } else { dp[sta|(1<<j)][j]+=dp[sta][i]; } } } } ans=(ans-m)/2; printf("%lld\n",ans); return 0; }
每天刷题,身体棒棒!
标签：contain cto scanf 一个 += 个数 不能 algo ati
原文地址：http://www.cnblogs.com/stxy-ferryman/p/7637383.html
