POJ3694 Network （无向图缩点+割边+并查集优化）

爱被打了一巴掌 2021-11-17 18:18 279阅读 0赞

## 题意：给出n个点，m条无向边的图，图中存在割边，问每加入一条新的边后的割边（桥）的数量。 ##

## 分析：《算法竞赛进阶指南》P407。 ##

## 代码： ##

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;const int SIZE = 200010;
    int head[SIZE], ver[SIZE * 2], Next[SIZE * 2];
    int dfn[SIZE], low[SIZE], c[SIZE];
    int n, m, t, tot, num, dcc, tc, T;
    bool bridge[SIZE * 2], v[SIZE];
    int hc[SIZE], vc[SIZE * 2], nc[SIZE * 2];
    int d[SIZE], f[SIZE][20], fa[SIZE];
    queue<int> q;
    
    void add(int x, int y) {
    	ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
    }
    
    void add_c(int x, int y) {
    	vc[++tc] = y, nc[tc] = hc[x], hc[x] = tc;
    }
    
    void tarjan(int x, int in_edge) {
    	dfn[x] = low[x] = ++num;
    	for (int i = head[x]; i; i = Next[i]) {
    		int y = ver[i];
    		if (!dfn[y]) {
    			tarjan(y, i);
    			low[x] = min(low[x], low[y]);
    			if (low[y] > dfn[x])
    				bridge[i] = bridge[i ^ 1] = true;
    		}
    		else if (i != (in_edge ^ 1))
    			low[x] = min(low[x], dfn[y]);
    	}
    }
    
    void dfs(int x) {
    	c[x] = dcc;
    	for (int i = head[x]; i; i = Next[i]) {
    		int y = ver[i];
    		if (c[y] || bridge[i]) continue;
    		dfs(y);
    	}
    }
    
    void bfs() {
    	q.push(1);
    	d[1] = 1;
    	while (q.size()) {
    		int x = q.front();
    		q.pop();
    		for (int i = hc[x]; i; i = nc[i]) {
    			int y = vc[i];
    			if (d[y]) continue;
    			d[y] = d[x] + 1;
    			f[y][0] = x;
    			for (int j = 1; j < 18; j++)
    				f[y][j] = f[f[y][j - 1]][j - 1];
    			q.push(y);
    		}
    	}
    }
    
    int lca(int x, int y) {
    	if (d[x] < d[y]) swap(x, y);
    	for (int i = 17; i >= 0; i--)
    		if (d[f[x][i]] >= d[y]) x = f[x][i];
    	if (x == y) return x;
    	for (int i = 17; i >= 0; i--)
    		if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    	return f[x][0];
    }
    
    int get(int x) {
    	if (x == fa[x]) return x;
    	return fa[x] = get(fa[x]);
    }
    
    int main() {
    	while (cin >> n >> m && n) {
    		// 多组数据，清零
    		tot = 1; num = dcc = 0;
    		for (int i = 1; i <= n; i++)
    			head[i] = dfn[i] = hc[i] = d[i] = c[i] = 0;
    		for (int i = 1; i <= 2 * m + 1; i++)
    			bridge[i] = 0;
    		for (int i = 1; i <= m; i++) {
    			int x, y;
    			scanf("%d%d", &x, &y);
    			add(x, y), add(y, x);
    		}
    		// 求割边、边双连通分量
    		for (int i = 1; i <= n; i++)
    			if (!dfn[i]) tarjan(i, 0);
    		for (int i = 1; i <= n; i++)
    			if (!c[i]) {
    				++dcc;
    				dfs(i);
    			}
    		// 缩点建图
    		tc = 1;
    		for (int i = 2; i <= tot; i++) {
    			int x = ver[i ^ 1], y = ver[i];
    			if (c[x] == c[y]) continue;
    			add_c(c[x], c[y]);
    		}
    		// 倍增lca预处理
    		bfs();
    		// 并查集初始化
    		for (int i = 1; i <= dcc; i++) fa[i] = i;
    		// 处理询问
    		cin >> t;
    		int ans = dcc - 1;
    		printf("Case %d:\n", ++T);
    		while (t--) {
    			int x, y;
    			scanf("%d%d", &x, &y);
    			x = c[x], y = c[y];
    			int p = lca(x, y);
    			x = get(x);
    			while (d[x] > d[p]) {
    				fa[x] = f[x][0];
    				ans--;
    				x = get(x);
    			}
    			y = get(y);
    			while (d[y] > d[p]) {
    				fa[y] = f[y][0];
    				ans--;
    				y = get(y);
    			}
    			printf("%d\n", ans);
    		}
    		cout << endl;
    	}
    }