UVALive 4262——Trip Planning——————【Tarjan 求强连通分量个数】

There is a road network comprised by M<tex2html_verbatim_mark> roads and N<tex2html_verbatim_mark> cities. For convenience, we use {1, 2,..., N}<tex2html_verbatim_mark> to denote the N<tex2html_verbatim_mark> cities. Each road between two cities i<tex2html_verbatim_mark> and j<tex2html_verbatim_mark> , where 1i<tex2html_verbatim_mark> , jN<tex2html_verbatim_mark> and ij<tex2html_verbatim_mark> , has two types: One type is bidirectional, which allows a citizen to drive a car from i<tex2html_verbatim_mark> to j<tex2html_verbatim_mark> (denoted by ij<tex2html_verbatim_mark> ) and from j<tex2html_verbatim_mark> to i<tex2html_verbatim_mark> (denoted by ji<tex2html_verbatim_mark> ). The other type is unidirectional, which allows a citizen to drive a car following exactly one direction, either from i<tex2html_verbatim_mark> to j<tex2html_verbatim_mark> or from j<tex2html_verbatim_mark> to i<tex2html_verbatim_mark> .

We say that City j<tex2html_verbatim_mark> is reachable from City i<tex2html_verbatim_mark> if one can drive a car from i<tex2html_verbatim_mark> to j<tex2html_verbatim_mark> , visiting a sequence of cities c₁, c₂,..., c_k<tex2html_verbatim_mark> for k 0<tex2html_verbatim_mark> , such thatic₁c₂...c_kj<tex2html_verbatim_mark> . (Every city is always reachable from itself.) A region is a maximal set of cities so that the following mutually reachable property holds: for two arbitrary cities i<tex2html_verbatim_mark> and j<tex2html_verbatim_mark> , i<tex2html_verbatim_mark> is reachable from j<tex2html_verbatim_mark> and j<tex2html_verbatim_mark> is also reachable from i<tex2html_verbatim_mark> . The adjective ``maximal" means that if we include any other city in the given region, the mutually reachable property cannot be retained. Given a road network, your task is to write a computer program to compute the number of regions in the road network.

Technical Specification

1. We use {1, 2,..., N}<tex2html_verbatim_mark> to denote the N<tex2html_verbatim_mark> cities.
2. M2000<tex2html_verbatim_mark> is a non-negative integer
3. N1000<tex2html_verbatim_mark> is a positive integer.
4. If a road between i<tex2html_verbatim_mark> and j<tex2html_verbatim_mark> is bidirectional, then we use two order pairs (i, j)<tex2html_verbatim_mark> and (j, i)<tex2html_verbatim_mark> to represent it. Otherwise, if a road between i<tex2html_verbatim_mark>and j<tex2html_verbatim_mark> is unidirectional from i<tex2html_verbatim_mark> to j<tex2html_verbatim_mark> (respectively, j<tex2html_verbatim_mark> to i<tex2html_verbatim_mark> ), we use ( i<tex2html_verbatim_mark> , j<tex2html_verbatim_mark> ) (respectively, ( j<tex2html_verbatim_mark> , i<tex2html_verbatim_mark> )) to represent it.

The input consists of a number of test cases. The first line of the input file contains an integer indicating the number of test cases to follow. Each test case consists of a road network, which has the following format: the first line of each test case contains two numbers, N<tex2html_verbatim_mark>and M<tex2html_verbatim_mark> , separated by a single space. The next M<tex2html_verbatim_mark> lines contain the description of M<tex2html_verbatim_mark> roads such that one line contains two cities representing an order pair (i, j)<tex2html_verbatim_mark> . Each line is represented by two positive numbers separated by a single space; the first number representing the former element in the order pair and the second number representing the latter element in the order pair. A ` 0‘ at the (M+ 2)<tex2html_verbatim_mark> -th line of each test case (except for the last test case) indicates the end of this test case.

The next test case starts after the previous ending symbol `0‘. Finally, a `-1‘ signals the end of the whole inputs.

The output contains one line for each test case. Each line contains an integer, which is the number of the regions in the given road network.

2 
3 2
1 2
1 3
0 
3 3
1 2
2 3
3 1
-1

3 
1



题目大意：给你n个点，m条有向边。问你这个图中的scc个数。

解题思路：求强连通分量的模板题，Tarjan算法水过。

/*
    Tarjan 
    求强连通分量个数
*/
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<stack>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 1e5+200;
vector<int>G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
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;
    while(!S.empty()) S.pop();
    memset(sccno , 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    for(int i = 1; i <= n; i++){
        if(!pre[i]) dfs(i);
    }
}
int main(){
    int T,n,m;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        int a,b;
        for(int i = 0; i < m; i++){
            scanf("%d%d",&a,&b);
            G[a].push_back(b);
        }
        scanf("%d",&a);
        find_scc(n);
        printf("%d\n",scc_cnt);
        for(int i = 0; i <= n; i++){
            G[i].clear();
        }
    }
    return 0;
}