CH24C 逃不掉的路

最新推荐文章于 2024-08-13 14:36:52 发布

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最新推荐文章于 2024-08-13 14:36:52 发布

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原文链接：http://www.cnblogs.com/CzxingcHen/p/9466486.html

本文提供了一个结合EDCC（一种图简化技术）与LCA（最近公共祖先查询）算法的实现示例，用于解决图论中寻找最近公共祖先的问题。通过EDCC将原图简化为凝聚图，再利用倍增法进行LCA查询，有效提升了算法效率。

edcc缩点之后跳倍增lca
丢个edcc缩点模板

Code:

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1e5 + 5;
const int M = 2e5 + 5;
const int Lg = 25;

int n, m, qn, tot = 1, head[N], dep[N], fa[N][Lg];
int dfsc = 0, dcc = 0, dfn[N], low[N], bel[N], ex[M], ey[M];
bool bri[M << 1], vis[N];

struct Edge {
    int to, nxt;
} e[M << 1];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

inline void read(int &X) {
    X = 0;
    char ch = 0;
    int op = 1;
    for(; ch > '9'|| ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline int min(int x, int y) {
    return x > y ? y : x;
}

void tarjan(int x, int inEdge) {
    dfn[x] = low[x] = ++dfsc;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if(dfn[x] < low[y]) bri[i] = bri[i ^ 1] = 1;
        } else if(i != (inEdge ^ 1))
            low[x] = min(low[x], dfn[y]);
    }
}

void dfs(int x) {
    bel[x] = dcc;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(bel[y] || bri[i]) continue;
        dfs(y);
    }
}

void dfs2(int x, int fat, int depth) {
    vis[x] = 1, fa[x][0] = fat, dep[x] = depth;
    for(int i = 1; i <= 20; i++)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(vis[y]) continue;
        dfs2(y, x, depth + 1);
    } 
}

void swap(int &x, int &y) {
    int t = x;
    x = y;
    y = t;
}

inline int getLca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 20; i >= 0; i--)
        if(dep[fa[x][i]] >= dep[y])
            x = fa[x][i];
    if(x == y) return x;
    for(int i = 20; i >= 0; i--)
        if(fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

int main() {
    read(n), read(m);
    for(int i = 1; i <= m; i++) {
        read(ex[i]), read(ey[i]);
        add(ex[i], ey[i]), add(ey[i], ex[i]);
    }
    
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i, 0);
    for(int i = 1; i <= n; i++)
        if(!bel[i]) {
            ++dcc;
            dfs(i);
        }
    
/*    for(int i = 1; i <= n; i++)
        printf("%d ", bel[i]);
    printf("\n");   */
    
    tot = 0;
    memset(head, 0, sizeof(head));
    for(int i = 1; i <= m; i++) {
        if(bel[ex[i]] == bel[ey[i]]) continue;
        add(bel[ex[i]], bel[ey[i]]), add(bel[ey[i]], bel[ex[i]]);
    }
    for(int i = 1; i <= n; i++)
        if(!vis[i]) dfs2(i, 0, 1);
    
    read(qn);
    for(int x, y; qn--; ) {
        read(x), read(y);
        x = bel[x], y = bel[y];
        if(x == y) puts("0");
        else printf("%d\n", dep[x] + dep[y] - 2 * dep[getLca(x, y)]);
    }
    return 0;
}