BZOJ3331 压力 （圆方树+树上差分）

题意

略

题解

求路径上的割点。

然后就直接圆方树上差分

CODE

#include <bits/stdc++.h>
using namespace std;
inline void rd(int &x) {
    char ch; for(;!isdigit(ch=getchar()););
    for(x=ch-'0';isdigit(ch=getchar());)x=x*10+ch-'0';
}
const int MAXN = 100005;
const int MAXM = 200005;
int n, m, q;
 
int fir[MAXN], to[MAXM<<1], nxt[MAXM<<1], cnt = 1;
inline void link(int u, int v) {
    to[++cnt] = v; nxt[cnt] = fir[u]; fir[u] = cnt;
    to[++cnt] = u; nxt[cnt] = fir[v]; fir[v] = cnt;
}
 
int dfn[MAXN], low[MAXN], tmr, stk[MAXN], top, tot;
 
vector<int>G[MAXN<<1];
 
void tarjan(int u, int ff) {
    stk[++top] = u;
    low[u] = dfn[u] = ++tmr;
    for(int i = fir[u], v; i; i = nxt[i])
        if((i^1) != ff) {
            if(!dfn[v=to[i]]) {
                tarjan(v, i);
                low[u] = min(low[u], low[v]);
                if(low[v] >= dfn[u]) {
                    ++tot;
                    do G[tot].push_back(stk[top]), G[stk[top]].push_back(tot);
                    while(stk[top--] != v);
                    G[tot].push_back(u), G[u].push_back(tot);
                }
            }
            else low[u] = min(low[u], dfn[v]);
        }
}
int dep[MAXN<<1], sz[MAXN<<1], tp[MAXN<<1], son[MAXN<<1], fa[MAXN<<1];
void dfs1(int u, int ff) {
    dep[u] = dep[fa[u]=ff] + (sz[u] = 1);
    for(int i = G[u].size()-1, v; i >= 0; --i)
        if((v=G[u][i]) != ff) {
            dfs1(v, u), sz[u] += sz[v];
            if(sz[v] > sz[son[u]]) son[u] = v;
        }
}
void dfs2(int u, int Tp) {
    tp[u] = Tp;
    if(son[u]) dfs2(son[u], Tp);
    for(int i = G[u].size()-1, v; i >= 0; --i)
        if((v=G[u][i]) != fa[u] && v != son[u])
            dfs2(v, v);
}
inline int lca(int u, int v) {
    while(tp[u] != tp[v]) {
        if(dep[tp[u]] > dep[tp[v]]) u = fa[tp[u]];
        else v = fa[tp[v]];
    }
    return dep[u] < dep[v] ? u : v;
}
int f[MAXN<<1];
bool vis[MAXN<<1];
int dfs(int u) {
    if(vis[u]) return f[u]; vis[u] = 1;
    for(int i = G[u].size()-1, v; i >= 0; --i)
        if((v=G[u][i]) != fa[u])
            f[u] += dfs(v);
    return f[u];
}
int main() {
    rd(n), rd(m), rd(q); tot = n;
    for(int i = 1, u, v; i <= m; ++i) rd(u), rd(v), link(u, v);
    for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i, 0), --top;
    dfs1(1, 0), dfs2(1, 1);
    for(int i = 1, u, v, Lca; i <= q; ++i) {
        rd(u), rd(v);
        Lca = lca(u, v);
        ++f[u], ++f[v], --f[Lca], --f[fa[Lca]];
    }
    for(int i = 1; i <= n; ++i) printf("%d
", dfs(i));
}

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附上圆方树学习链接：
租酥雨的博客
YoungNeal的博客