山东省第二届aThe Largest SCC（强连通分量）

The Largest SCC

1000 ms 65536 KiB
Accepted/Submissions: 3/8 (37.50%)

Description

Consider a directed graph with N (1 <= N <= 1000) vertices and M (0 <= M <= 20000) edges. The edges are numbered from 1 to M and the vertices are numbered from 1 to N. Now I will make ONE edge bidirectional, and you are to tell me the number of vertices of the largest strong connected components in the new graph.The largest strong connected components is the strong connected components which has the most vertices. After the operation, I will change the edge back. There will be up to Q (1 <= Q <= 20000) such queries.

Input

At the firest of the input comes an integer t, indicates the testcases to follow. The first line of each case contains three numbers N, M and Q. Then there will be M lines, each of them contains two numbers a,b (a! = b; 1 <= a; b <= N) means there is a directed edge between a and b. The last of each case contains Q lines, each of them contains one integer q, means the edge numbered q will be change to bidirectional. There will not be duplicated edges.

Output

For every query, output one line contains only one integer number, which is the number of vertices of the biggest strong connected components.

Sample

Input

Output

Copy2
3

需要注意的是之前数据的保存和恢复。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 50005;
const int maxm = 50005;
const int inf = 0x3f3f3f3f;
int tmaxn, remaxn;
struct edge {
	int u, v, ne;
}ed[maxm];
int head[maxn], cnt;
int low[maxn], dfn[maxn], stack[maxn], belong[maxn];//belong存的是1到n的归属
int indexn, top; int scc;//强连通分量的个数；
int flag = 0;
bool instack[maxn]; int sum[maxn];
void add(int u, int v) {
	ed[cnt].v = v; ed[cnt].ne = head[u];
	ed[cnt].u = u;head[u] = cnt++;
}
void tarjan(int u) {
//	cout << u << endl;
	int v;
	low[u] = dfn[u] = ++indexn;
	stack[top++] = u;
	instack[u] = 1;
	for (int s = head[u]; ~s; s = ed[s].ne) {
		v = ed[s].v;
		if (!dfn[v]) {
		//	cout << u << " go " << v << endl;
			tarjan(v);
			if (low[u] > low[v])low[u] = low[v];
		}
		else if (instack[v] && low[u] > dfn[v])
			low[u] = dfn[v];
	}
	if (low[u] == dfn[u]) {
		scc++;
		do {
			v = stack[--top];
			instack[v] = 0;
			if (!flag)
				belong[v] = scc;
			sum[scc]++;
			tmaxn = max(tmaxn, sum[scc]);
		} while (v != u);
	}
}
void solve(int n) {
	memset(dfn, 0, sizeof(dfn));
	memset(instack, 0, sizeof(instack));
	memset(sum, 0, sizeof(sum));
	indexn = scc = top = 0;
	for (int s = 1; s <= n; s++)
		if (!dfn[s])
			tarjan(s);
}
void init() {
	cnt = 0;
	memset(head, -1, sizeof(head));
}
int main() {
	int te;
	ios::sync_with_stdio(0);
	cin >> te;
	while (te--) {
		flag = 0;
		int n, m, q;
		cin >> n >> m >> q;
		init();
		while (m--) {
			int a, b;
			cin >> a >> b;
			add(a, b);
		}	
		tmaxn = 0, remaxn;
		solve(n);
		flag = 1;
		remaxn = tmaxn;
		while (q--) {
			int a; cin >> a; a--;
			int u = ed[a].v, v = ed[a].u;
			if (belong[u] == belong[v]) {
				cout << tmaxn << "\n";
				continue;
			}
			add(u, v);
			indexn = top = 0;
			memset(dfn, 0, sizeof(dfn));
			tarjan(v);
			cout << tmaxn << "\n";
			tmaxn = remaxn;
			head[u] = ed[--cnt].ne;
		}
	}
	return 0;
}