按照vdcc缩点之后一条边只会属于一个新的点集,由于这棵树上满足(不是割点) - (割点) - (不是割点)的连接方法,所以求两条边之间的必经点就是(树上距离 / 2),倍增跳lca即可
考虑到缩点后树上点数的编号可能超过n,所以与树有关的数组开两倍N
又是一个模板

Code:

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std; const int N = 1e4 + ;
const int M = 1e5 + ;
const int Lg = ; int n, m, qn, tot, head[][N << ], top, sta[N];
int dfsc, dfn[N], low[N], dccCnt, root;
int bel[N], eb[M], fa[N << ][Lg], dep[N << ], id[N];
bool cut[N], vis[N << ];
vector <int> dcc[N]; struct Edge {
int to, nxt, id;
} e[M << ]; inline void add(int type, int from, int to, int eid) {
e[++tot].to = to;
e[tot].id = eid;
e[tot].nxt = head[type][from];
head[type][from] = tot;
} inline void read(int &X) {
X = ;
char ch = ;
int op = ;
for(; ch > ''|| ch < ''; ch = getchar())
if(ch == '-') op = -;
for(; ch >= '' && ch <= ''; ch = getchar())
X = (X << ) + (X << ) + ch - ;
X *= op;
} inline int min(int x, int y) {
return x > y ? y : x;
} void tarjan(int x) {
dfn[x] = low[x] = ++dfsc;
sta[++top] = x;
if(x == root && head[][x] == ) {
dcc[++dccCnt].push_back(x);
return;
} int son = ;
for(int i = head[][x]; i; i = e[i].nxt) {
int y = e[i].to;
if(!dfn[y]) {
tarjan(y);
low[x] = min(low[x], low[y]);
if(dfn[x] <= low[y]) {
++son;
if(x != root || son > ) cut[x] = ;
++dccCnt;
for(int z; ; ) {
z = sta[top--];
dcc[dccCnt].push_back(z);
if(z == y) break;
}
dcc[dccCnt].push_back(x);
}
} else low[x] = min(low[x], dfn[y]);
}
} void dfs(int x, int fat, int depth) {
vis[x] = , fa[x][] = fat, dep[x] = depth;
for(int i = ; i <= ; i++)
fa[x][i] = fa[fa[x][i - ]][i - ];
for(int i = head[][x]; i; i = e[i].nxt) {
int y = e[i].to;
if(vis[y]) continue;
dfs(y, x, depth + );
}
} inline 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 = ; i >= ; i--)
if(dep[fa[x][i]] >= dep[y])
x = fa[x][i];
if(x == y) return x;
for(int i = ; i >= ; i--)
if(fa[x][i] != fa[y][i])
x = fa[x][i], y = fa[y][i];
return fa[x][];
} int main() {
for(; ; ) {
read(n), read(m);
if(n == && m == ) break;
tot = ;
memset(head, , sizeof(head));
for(int x, y, i = ; i <= m; i++) {
read(x), read(y);
add(, x, y, i), add(, y, x, i);
} dfsc = dccCnt = top = ;
memset(dfn, , sizeof(dfn));
memset(low, , sizeof(low));
memset(cut, , sizeof(cut));
for(int i = ; i <= n; i++) dcc[i].clear(); for(int i = ; i <= n; i++)
if(!dfn[i]) tarjan(root = i); int now = dccCnt;
for(int i = ; i <= n; i++)
if(cut[i]) id[i] = ++now; /* printf("\n");
for(int i = 1; i <= dccCnt; i++, printf("\n")) {
for(unsigned int j = 0; j < dcc[i].size(); j++)
printf("%d ", dcc[i][j]);
}
printf("\n"); */ for(int i = ; i <= dccCnt; i++) {
for(unsigned int j = ; j < dcc[i].size(); j++) {
int x = dcc[i][j];
if(cut[x]) add(, id[x], i, ), add(, i, id[x], );
bel[x] = i;
} for(unsigned int j = ; j < dcc[i].size(); j++) {
int x = dcc[i][j];
for(int k = head[][x]; k; k = e[k].nxt) {
int y = e[k].to;
if(bel[y] == i) eb[e[k].id] = i;
}
}
} /* for(int i = 1; i <= n; i++)
printf("%d ", bel[i]);
printf("\n"); for(int i = 1; i <= m; i++)
printf("%d ", eb[i]);
printf("\n"); */ memset(dep, , sizeof(dep));
memset(fa, , sizeof(fa));
memset(vis, , sizeof(vis));
for(int i = ; i <= now; i++)
if(!vis[i]) dfs(i, , ); read(qn);
for(int x, y; qn--; ) {
read(x), read(y);
x = eb[x], y = eb[y];
if(x == y) puts("");
else printf("%d\n", (dep[x] + dep[y] - * dep[getLca(x, y)]) / );
}
}
return ;
}

lyd给的std好难看

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