标签:删除 namespace 没有 getc 赋值 -- add dig print
题目大意:多组数据,每组数据给一张图,多组询问,每个询问给一个点集,要求删除一个点,使得至少点集中的两个点互不连通,输出方案数
题解:圆方树,发现使得两个点不连通的方案数就是它们路径上的原点个数。如何处理重复?可以按圆方树的$dfn$序排序,相邻两点求一下贡献,这样贡献就被重复计算了两次,除去$k$个询问点就行了。还有每次计算中$lca$没有被统计,发现排序后第一个点和最后一个点的$lca$一定是深度最浅的,所以只有这个点没有被统计答案,加上即可
卡点:1.圆方树$dfn$数组没赋值
2.$LCA$的$log$太小
C++ Code:
#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 200010
#define maxm 200010
int Tim, n, m, LCA;
inline int min(int a, int b) {return a < b ? a : b;}
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
struct Tree {
#define root 1
#define fa(u) dad[u][0]
#define M 18
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxm << 1];
inline void addE(int a, int b) {e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;}
inline void add(int a, int b) {
addE(a, b);
addE(b, a);
}
int dad[maxn][M], dep[maxn], sz[maxn];
int dfn[maxn], idx;
void dfs(int u = root) {
dfn[u] = ++idx;
for (int i = 1; i < M; i++) dad[u][i] = dad[dad[u][i - 1]][i - 1];
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa(u)) {
sz[v] = sz[u] + int(v <= n);
dep[v] = dep[u] + 1;
fa(v) = u;
dfs(v);
}
}
}
inline int LCA(int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
for (int i = dep[x] - dep[y]; i; i &= i - 1) x = dad[x][__builtin_ctz(i)];
if (x == y) return x;
for (int i = M - 1; ~i; i--) if (dad[x][i] != dad[y][i]) x = dad[x][i], y = dad[y][i];
return fa(x);
}
inline int len(int x, int y) {
return sz[x] + sz[y] - (sz[::LCA = LCA(x, y)] << 1);
}
inline void init() {
memset(head, 0, sizeof head); cnt = 0;
memset(dfn, 0, sizeof dfn); idx = 0;
sz[root] = 0;
}
#undef root
#undef fa
#undef M
} T;
struct Graph {
#define root 1
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxm << 1];
inline void addE(int a, int b) {e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;}
inline void add(int a, int b) {
addE(a, b);
addE(b, a);
}
int DFN[maxn], low[maxn], idx, CNT;
int S[maxn], top, tmp;
void tarjan(int u = root) {
DFN[u] = low[u] = ++idx;
S[++top] = u;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!DFN[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
if (low[v] >= DFN[u]) {
CNT++;
T.add(CNT, u);
do {
T.add(CNT, tmp = S[top--]);
} while (tmp != v);
}
} else low[u] = min(low[u], DFN[v]);
}
}
inline void init(int n) {
memset(head, 0, sizeof head); cnt = 0;
memset(DFN, 0, sizeof DFN); idx = 0;
CNT = n;
}
#undef root
} G;
#define Online_Judge
#define read() R::READ()
#include <cctype>
namespace R {
int x;
#ifdef Online_Judge
char *ch, op[1 << 26];
inline void init() {
fread(ch = op, 1, 1 << 26, stdin);
}
inline int READ() {
while (isspace(*ch)) ch++;
for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15);
return x;
}
#else
char ch;
inline int READ() {
ch = getchar();
while (isspace(ch)) ch = getchar();
for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
return x;
}
#endif
}
int s[maxn];
inline bool cmp(int a, int b) {return T.dfn[a] < T.dfn[b];}
int main() {
#ifdef Online_Judge
R::init();
#endif
Tim = read();
while (Tim --> 0) {
G.init(n = read()), T.init();
for (int i = m = read(); i; i--) G.add(read(), read());
G.tarjan();
T.dfs();
int Q = read();
while (Q --> 0) {
int k = read(), ans = 0;
for (int i = 0; i < k; i++) s[i] = read();
std::sort(s, s + k, cmp);
s[k] = s[0];
for (int i = 0; i < k; i++) ans += T.len(s[i], s[i + 1]);
printf("%d\n", (ans >> 1) - k + int(LCA <= n));
}
}
return 0;
}
标签:删除 namespace 没有 getc 赋值 -- add dig print
原文地址:https://www.cnblogs.com/Memory-of-winter/p/9650984.html