HDU 3836 Equivalent Sets(强连通分量)
2016-02-04 17:44
399 查看
思路:和HDU2726基本一样
Description
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
Input
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Output
For each case, output a single integer: the minimum steps needed.
Sample Input
Sample Output
Hint
#include <cstdio> #include <queue> #include <cstring> #include <iostream> #include <cstdlib> #include <algorithm> #include <vector> #include <map> #include <string> #include <set> #include <ctime> #include <cmath> #include <cctype> #include <stack> using namespace std; #define maxn 20000+100 #define LL long long int cas=1,T; vector<int>G[maxn]; int pre[maxn]; int lowlink[maxn]; int sccno[maxn]; int dfs_clock,scc_cnt; int n,m; stack<int>S; void dfs(int u) { pre[u]=lowlink[u]=++dfs_clock; S.push(u); for (int i = 0;i<G[u].size();i++) { int v = G[u][i]; if (!pre[v]) { dfs(v); lowlink[u] = min(lowlink[u],lowlink[v]); } else if (!sccno[v]) { lowlink[u] = min (lowlink[u],pre[v]); } } if (lowlink[u] == pre[u]) { scc_cnt++; for (;;) { int x = S.top();S.pop(); sccno[x] = scc_cnt; if (x==u) break; } } } void find_scc(int n) { dfs_clock=scc_cnt=0; memset(sccno,0,sizeof(sccno)); memset(pre,0,sizeof(pre)); for (int i = 1;i<=n;i++) if (!pre[i]) dfs(i); } int in[maxn]; int out[maxn]; int main() { //freopen("in","r",stdin); while(scanf("%d%d",&n,&m)!=EOF){ for (int i = 0;i<=n;i++) G[i].clear(); memset(in,0,sizeof(in)); memset(out,0,sizeof(out)); for (int i = 1;i<=m;i++) { int u,v; scanf("%d%d",&u,&v); G[u].push_back(v); } find_scc(n); for (int i = 1;i<=scc_cnt;i++) { in[i]=out[i]=1; } for (int u = 1;u<=n;u++) for (int i =0;i<G[u].size();i++) { int v = G[u][i]; if (sccno[u] != sccno[v]) in[sccno[v]]=out[sccno[u]]=0; } int a=0,b=0; for (int i = 1;i<=scc_cnt;i++) { if (in[i]) a++; if (out[i]) b++; } if (scc_cnt !=1) printf("%d\n",max(a,b)); else printf("0\n"); } return 0; }
Description
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
Input
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Output
For each case, output a single integer: the minimum steps needed.
Sample Input
4 0 3 2 1 2 1 3
Sample Output
4 2
Hint
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
相关文章推荐
- 【第一行代码】Android中UI总结
- HUE中通过SQL进行数据分析
- getContextPath、getServletPath、getRequestURI的区别
- request请求doGet处理中文乱码问题
- NSFetchedResultsController 在CoreData中的使用 配合UITableView
- HDU 2767 Proving Equivalences(强连通分量)
- SVN检出项目后报build.properties does not exit完整版解决方案
- 网上图书商城项目学习笔记-035工具类之JdbcUtils及TxQueryRunner及C3P0配置
- 不无聊的序列(Non-boring sequences,Bzoj4059,CERC2012)
- hue忘记管理员登陆密码
- UVA1626 - Brackets sequence
- UGUI中带参数事件的使用,itween动画注意事项以及切换场景
- 内核request_mem_region 和 ioremap的理解
- ServletRequest获取不到getParameter和filter获取不到request请求中的Parameter
- 【POJ2478】Farey Seque
- GitHub——Pull Request
- IOS 颜色值#ffffff转UIColor
- Android 动画 ValueAnimator(四)
- 将数据库查询到的ueditor内容现在另外一个ueditor中
- CustomSliderView