POJ3694 Network

Description

给定$N$个点和 $M$条边的无向联通图， 有$Q$ 次操作， 连接两个点的边， 问每次操作后的图中有几个桥

Solution

首先Tarjan找出边双联通分量， 每个双联通分量缩成一个点， 就构成了一棵树， 每一条树边都是桥。

执行连$u, v$ 边时， 用并查集跳到没有桥的深度最浅并且深度比$lca$深的点， 将它与父节点的并查集合并， 再接着跳。

每跳一次， 桥的数量就减少$1$。

另外感谢Iowa 神犇提醒我$cut$数组要开$M << 1$, 不是 $N << 1$, 拯救了$RE$崩溃的我呜呜

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rd read()
using namespace std;

const int N = 1e5 + 1e4;
const int M = 2e5 + 1e4;

int n, m, dfn[N], low[N], cnt; 
int head[N], tot;
int Head[N], Tot;
int col[N], col_num, father[N], ans;
int top[N], son[N], size[N], f[N], dep[N];
int cut[M << 1];

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

int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

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

void Add(int u, int v) {
    E[++Tot].to = v;
    E[Tot].nxt = Head[u];
    Head[u] = Tot;
}

void dfs1(int u) {
    size[u] = 1;
    for(int i = Head[u]; i; i = E[i].nxt) {
        int nt = E[i].to;
        if(nt == f[u]) continue;
        f[nt] = u;
        dep[nt] = dep[u] + 1;
        dfs1(nt);
        size[u] += size[nt];
        if(size[nt] > size[son[u]]) son[u] = nt;
    }
}

void dfs2(int u) {
    if(!son[u]) return;
    top[son[u]] = top[u];
    dfs2(son[u]);
    for(int i = Head[u]; i; i = E[i].nxt) {
        int nt = E[i].to;
        if(nt == f[u] || nt == son[u]) continue;
        top[nt] = nt;
        dfs2(nt);
    }
}

int LCA(int x, int y) {
    for(; top[x] != top[y];) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        x = f[top[x]];
    }
    if(dep[x] < dep[y]) swap(x, y);
    return y;
}

int get(int x) {
    return father[x] == x? x : father[x] = get(father[x]);
}

void merg(int x, int y) {
    x = get(x); y = get(y);
    father[x] = y;
}

int ch(int x) {
    return ((x + 1) ^ 1) - 1;
}

void tarjan(int u, int pre) {
    dfn[u] = low[u] = ++cnt;
    for(int i = head[u]; i; i = e[i].nxt) {
        if(i == ch(pre)) continue; 
        int nt = e[i].to;
                if(!dfn[nt]) {
                    tarjan(nt, i);
                    low[u] = min(low[u], low[nt]);
                    if(low[nt] > dfn[u]) {
                        cut[ch(i)] = cut[i] = 1;
                        ans++;
                    }
                }
                else low[u] = min(low[u], dfn[nt]);
    }
}

void dfs(int u) {
    col[u] = col_num;
    for(int i = head[u]; i; i = e[i].nxt) {
        int nt = e[i].to;
        if(col[nt] || cut[i]) continue;
        dfs(nt);
        //blo[col_num].push_back(nt);
    }
}

void work() {
    ans = Tot = tot = cnt = col_num = 0;
    memset(Head, 0, sizeof(Head));
    memset(head, 0, sizeof(head));
    memset(cut, 0, sizeof(cut));
    memset(dfn, 0, sizeof(dfn));
    memset(col, 0, sizeof(col));
    memset(son, 0, sizeof(son));
    memset(size, 0, sizeof(size));
    for(int i = 1; i <= m; ++i) {
        int u = rd, v = rd;
        add(u, v); add(v, u);
    }
    for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i, 0);
    for(int i = 1; i <= n; ++i) if(!col[i]) {
        ++col_num; dfs(i);
    }
    for(int i = 1; i <= tot; ++i) {
        int x = e[i].to, y = e[ch(i)].to;
        if(col[x] == col[y]) continue;
        Add(col[x], col[y]);
    }
    for(int i = 1; i <= col_num; ++i) father[i] = i;
    dfs1(1);
    top[1] = 1; dfs2(1);
    int T = rd;
    for(; T; T--) {
        int u = col[rd], v = col[rd], lca = LCA(u, v);
        u = get(u); v = get(v);
        while(dep[u] > dep[lca]) {
            merg(u, f[u]);
            u = get(u);
            ans --;
        }
        while(dep[v] > dep[lca]) {
            merg(v, f[v]);
            v = get(v);
            ans --;
        }
        printf("%d
", ans);
    }
}

int main()
{
    for(int i = 1; ; ++i) {
        n = rd; m = rd;
        if(!n && !m) return 0;
        printf("Case %d:
", i);
        work();
        putchar('
');
    }
}