題目描述
不久之前,Mirko建立了一個旅行社,名叫“極地之夢”。這家旅行社在北極附近購買了
Mirko的旅行社遭受一次重大打擊,以至於觀光遊輪已經不划算了。旅行社將在冰島之間建造大橋,並用觀光巴士來運載遊客。Mirko希望開發一個電腦程序來管理這些大橋的建造過程,以免有不可預料的錯誤發生。
這些冰島從
你的程序需要處理以下三種命令:
輸入格式
第一行一個正整數
第二行
第三行一個正整數
接下來
輸出格式
對於每個bridge命令與excursion命令,輸出一行,爲題目描述所示。
一開始以爲是LCT,然而我太弱以至於根本不會LCT,DJY提醒我只要樹剖就可以了,強行被報標算,不A都對不起良心了。對於bridge只要用並查集判斷即可,對於excursion在第一遍並查集的時候順便記一下要不要輸
#include<bits/stdc++.h>
const int N = 3e4 + 10;
const int M = 1e5 + 10;
template <typename T> void read(T &x) {
x = 0; char c = getchar();
for (; !isdigit(c); c = getchar());
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
}
int n, m, a[N], f[N], first[N], s;
char op[10];
struct edge{
int y, next;
}mp[N * 4];
void ins(int x, int y) {
mp[++s] = (edge) {y, first[x]}; first[x] = s;
mp[++s] = (edge) {x, first[y]}; first[y] = s;
}
int find(int x) {
if (f[x] == x) return x;
return f[x] = find(f[x]);
}
namespace force {
int x, y;
void bridge(int x, int y) {
int f1 = find(x);
int f2 = find(y);
if (f1 == f2) {printf("no\n"); return;}
printf("yes\n");
f[f1] = f2;
ins(x, y);
}
bool dfs(int x, int y, int sum, int last) {
if (x == y) {printf("%d\n", sum); return 1;}
for (int t=first[x]; t; t=mp[t].next)
if (mp[t].y != last && dfs(mp[t].y, y, sum + a[mp[t].y], x))
return 1;
return 0;
}
int main() {
read(n);
for (int i=1; i <= n; i++)
read(a[i]), f[i] = i;
read(m);
while (m--) {
scanf("%s", op);
read(x); read(y);
if (op[0] == 'b') bridge(x, y);
if (op[0] == 'p') a[x] = y;
if (op[0] == 'e') if (!dfs(x, y, a[x], -1)) printf("impossible\n");
}
return 0;
}
}
namespace try_it {
int x[M], y[M];
char opt[M];
int fa[N][20], size[N], top[N], dep[N], pos[N], cnt;
int lc[N * 4], rc[N * 4], sum[N * 4], node;
bool z[M], vis[N];
bool bridge(int x, int y) {
int f1 = find(x);
int f2 = find(y);
if (f1 == f2) return 0;
f[f1] = f2; ins(x, y);
return 1;
}
void dfs1(int x) {
vis[x] = 1;
for (int t=first[x]; t; t=mp[t].next)
if (!vis[mp[t].y]) dfs1(mp[t].y);
}
int dfs2(int x) {
for (int i=1; dep[x] - (1 << i) >= 0; i++)
fa[x][i] = fa[fa[x][i - 1]][i - 1];
size[x] = 1;
for (int t=first[x]; t; t=mp[t].next)
if (mp[t].y != fa[x][0])
fa[mp[t].y][0] = x,
dep[mp[t].y] = dep[x] + 1,
size[x] += dfs2(mp[t].y);
return size[x];
}
void dfs3(int x, int cap) {
top[x] = cap;
pos[x] = ++cnt;
int mxszi = 0;
for (int t=first[x]; t; t=mp[t].next)
if (mp[t].y != fa[x][0] && size[mp[t].y] > size[mxszi])
mxszi = mp[t].y;
if (!mxszi) return;
dfs3(mxszi, cap);
for (int t=first[x]; t; t=mp[t].next)
if (mp[t].y != fa[x][0] && mp[t].y != mxszi)
dfs3(mp[t].y, mp[t].y);
}
void build(int i, int l, int r) {
if (l == r) return;
lc[i] = ++node; build(lc[i], l, (l + r) / 2);
rc[i] = ++node; build(rc[i], (l + r) / 2 + 1, r);
}
void chg(int i, int l, int r, int x, int c) {
if (l == r) {sum[i] = c; return;}
if (x <= (l + r) / 2)
chg(lc[i], l, (l + r) / 2, x, c);
else
chg(rc[i], (l + r) / 2 + 1, r, x, c);
sum[i] = sum[lc[i]] + sum[rc[i]];
}
int lca(int x, int y) {
if (dep[x] < dep[y]) std::swap(x, y);
for (int i=19; i >= 0; i--)
if (dep[x] - (1 << i) >= dep[y])
x = fa[x][i];
for (int i=19; i >= 0; i--)
if (fa[x][i] != fa[y][i])
x = fa[x][i],
y = fa[y][i];
if (x == y) return x;
return fa[x][0];
}
int qry(int i, int l, int r, int _l, int _r) {
if (l > _r || r < _l) return 0;
if (l >= _l && r <= _r) return sum[i];
return qry(lc[i], l, (l + r) / 2, _l, _r) + qry(rc[i], (l + r) / 2 + 1, r, _l, _r);
}
int qry(int x, int y) {
int l = lca(x, y), ret = - a[l];
while (dep[top[x]] > dep[l]) ret += qry(1, 1, cnt, pos[top[x]], pos[x]), x = fa[top[x]][0];
ret += qry(1, 1, cnt, pos[l], pos[x]);
while (dep[top[y]] > dep[l]) ret += qry(1, 1, cnt, pos[top[y]], pos[y]), y = fa[top[y]][0];
ret += qry(1, 1, cnt, pos[l], pos[y]);
return ret;
}
int main() {
read(n);
for (int i=1; i <= n; i++)
read(a[i]), f[i] = i;
read(m);
for (int i=1; i <= m; i++) {
scanf("%s", op);
opt[i] = op[0];
read(x[i]); read(y[i]);
if (opt[i] == 'b') z[i] = bridge(x[i], y[i]);
if (opt[i] == 'e') z[i] = find(x[i]) == find(y[i]);
}
for (int i=1; i <= n; i++)
if (!vis[i]) dfs1(i), ins(n + 1, i);
dfs2(n + 1);
dfs3(n + 1, n + 1);
node = 1; build(1, 1, cnt);
for (int i=1; i <= n; i++) chg(1, 1, cnt, pos[i], a[i]);
for (int i=1; i <= m; i++) {
if (opt[i] == 'b') if (z[i]) printf("yes\n"); else printf("no\n"); else;
if (opt[i] == 'p') chg(1, 1, cnt, pos[x[i]], y[i]), a[x[i]] = y[i];
if (opt[i] == 'e') if (z[i]) printf("%d\n", qry(x[i], y[i]));
else printf("impossible\n"); else;
}
return 0;
}
}
int main() {
//force::main();
try_it::main();
return 0;
}