Mining Your Own Business LA 5135
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题目:https://vjudge.net/problem/UVALive-5135
题目大意
地下有几个点被n条隧道连接,让你在某个点设置太平井使的不管取消任意一个点,其他点都能到达存在太平井的点。问最少要安装几个太平井,安装的方案数。
分析
不愧是WF的题就是难的不要不要的,但是考点还是可以练习下的,抱着A了就是赚了的心态学习了下这道题。
这个题的模型是, 在一个无向图上选择任一少的点涂黑,使得任一删除一个点后,每个联通分量至少有一个黑点。那么根据BCC有:
- 不能涂割点。
- 一个BCC涂一个点
进一步可以发现,当一个BCC只有一个割点时,任选一个非割点涂即可。
当整个图没有割点时,只用图两个点,容易想到方案数为 C(n, 2).
代码
/******************************************************************** * File Name: Mining_Your_Own_Business.cpp * Author: Sequin * mail: Catherine199787@outlook.com * Created Time: 五 9/22 20:05:59 2017 *************************************************************************/#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <ctype.h>#include <set>#include <vector>#include <cmath>#include <bitset>#include <algorithm>#include <climits>#include <string>#include <list>#include <cctype>#include <cstdlib>#include <fstream>#include <sstream>using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define pi acos(-1.0)#define pii pair<int,int>#define ll long long#define MAX 10000005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)#define maxn 100005ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}struct Edge{ int u, v;};int pre[maxn], iscut[maxn], bccno[maxn], dfs_clock, bcc_cnt;vector<int> G[maxn], bcc[maxn];stack<Edge> S;int dfs(int u, int fa) { int lowu = pre[u] = ++dfs_clock; int child = 0; for(int i = 0; i < G[u].size(); i++) { int v = G[u][i]; Edge e = (Edge) {u, v}; if(!pre[v]) { S.push(e); child++; int lowv = dfs(v, u); lowu = min(lowv, lowu); if(lowv >= pre[u]) { bcc_cnt++; bcc[bcc_cnt].clear(); iscut[u] = true; W(1) { Edge x = S.top(); S.pop(); if(bccno[x.u] != bcc_cnt) { bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt; } if(bccno[x.v] != bcc_cnt) { bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt; } if(x.u == u && x.v == v){ break; } } } } else if(pre[v] < pre[u] && v != fa) { S.push(e); lowu = min(pre[v], lowu); } } if(fa < 0 && child == 1) { iscut[u] = 0; } return lowu;}void find_bcc(int N) { MEM(pre, 0); MEM(iscut, 0); MEM(bccno, 0); dfs_clock = 0; bcc_cnt = 0; for(int i = 1; i <= N; i++) { if(!pre[i]){ dfs(i, -1); } }}int mm[100005][2];int main() { std::ios::sync_with_stdio(false); int n; int T = 0; while(cin >> n && n) { T++; int maxnum = 0; MEM(mm, 0); for(int i = 0; i < n; i++) { int a, b; cin >> a >> b; mm[i][0] = a; mm[i][1] = b; a = max(a, b); maxnum = max(a, maxnum); } for(int i = 1; i <= maxnum; i++) { G[i].clear(); } for(int i = 0; i < n; i++){ int a = mm[i][0]; int b = mm[i][1]; G[a].push_back(b); G[b].push_back(a); } find_bcc(maxnum); ll ret = 0; ll ans = 1; for(int i = 1; i <= bcc_cnt; i++) { int cut_cnt = 0; for(int j = 0; j < bcc[i].size(); j++) { if(iscut[bcc[i][j]]) { cut_cnt++; } } if(cut_cnt == 1) { ret++; ans *= (ll)(bcc[i].size() - cut_cnt); } } if(bcc_cnt == 1) { ret = 2; ans = maxnum * (maxnum - 1) / 2; } cout << "Case " << T << ": "; cout << ret << " " << ans << endl; }}
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