Network POJ - 3694（边双连通分量， LCA）

最新推荐文章于 2020-06-21 21:56:29 发布

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最新推荐文章于 2020-06-21 21:56:29 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43326028/article/details/95469188

博客围绕网络管理员管理大型网络展开，该网络由计算机和连接组成，部分连接为桥，管理员计划逐个添加新连接消除桥。给出输入输出格式及示例，还提及通过回溯更新low值、标记桥，利用节点深度和父节点找LCA来计算桥数量。

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A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can’t be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.

You are to help the administrator by reporting the number of bridges in the network after each new link is added.

Input
The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).
Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.
The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.
The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.

The last test case is followed by a line containing two zeros.

Output
For each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.

Sample Input
3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
Sample Output
Case 1:
1
0

Case 2:
2
0

比如以下一副图，一开始各点的dfn值为圈中的值，low值为旁边绿色的值。
在这里插入图片描述
从7连到3时，开始回溯，更新low值

下图中紫色的点都被标记为1，记录了桥，如果查询图中橙色的两个点x, y，左边的点x的dfn值深，一步步找它的父节点，更新x=f[x], 当x, y的深度一样后，一起往上找父节点，由于下图中，a[x], a[y]都等于1，所以桥的数量减去了2，x=f[x],y=f[y]，此时x=y,找到LCA

在这里插入图片描述

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#define LL long long 
#define mem(a, b) memset(a, b, sizeof(a))
#define N 500010 
#define MOD
using namespace std;

int dfn[N], low[N], num;
int n, m, Q, ans; 
int f[N];
int a[N];
int ver[N*2], Next[2*N], head[N], tot;

void add(int x, int y) {//邻接表 
	ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
} 

void init() {
	tot = 0, num = 0, ans = 0;
	mem(head, -1);
	mem(low, 0);
	mem(dfn, 0);
	for(int i = 1; i<=n; i++) {
		f[i] = i;
	} 
	mem(a, 0);
}

void tarjan(int x, int root) { 
	low[x] = ++num;
	dfn[x] = dfn[root] + 1;
	for (int i = head[x]; i!=-1; i=Next[i]) {
		int y=ver[i];
		if(!dfn[y]) {
			f[y] = x;
			tarjan(y, x);
			low[x] = min(low[x], low[y]);
			if(low[y] > dfn[x]) {
				ans++;
				a[y] = 1;
			}
		}
		else if(root != y)
			low[x] = min(low[x], dfn[y]);
	}
}

void lca(int x, int y) {
	while(dfn[x] < dfn[y]) {
		if(a[y]) {
			ans--;
			a[y] = 0;
		}
		y = f[y];
	}
	while(dfn[x] > dfn[y]) {
		if(a[x]) {
			ans--;
			a[x] = 0;
		}
		x = f[x];
	}
	while(x != y) {
		if(a[x]) {
			ans--;
			a[x] = 0;
		}
		if(a[y]) {
			ans--;
			a[y] = 0;
		}
		x=f[x], y=f[y];
	}
}



int main() {
	int iCase = 0;
	while(scanf("%d%d", &n, &m)!=EOF) {
		if(n == 0 && m == 0) break;
		init();
		int x, y;
		for(int i = 1; i <= m; i++) {
			
			scanf("%d%d", &x, &y);
			add(x, y), add(y, x);
		}
		tarjan(1,0);		
		scanf("%d", &Q);
		printf("Case %d:\n",++iCase);
		for(int i = 1; i <= Q; i++) {
			scanf("%d%d",&x, &y);
			lca(x, y);
			printf("%d\n",ans);
		}
	}
}