Description
Solution
可以当作一种经典案例
给定数字后乘积如何最大
Code
#include<bits/stdc++.h>
using namespace std;
#define N 800010
struct rec {
int nxt, ver, val;
} t[N];
struct Rec {
int u, v, w;
} e[N];
int n, m, q, cnt, tot, u, v;
int fa[N], head[N], deep[N], f[N][21], h[N][21];
inline int read() {
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
int find(int x) {
if (fa[x] == x) return x;
return fa[x] = find(fa[x]);
}
inline void Merge(int x, int y) {
fa[find(x)] = find(y);
}
inline bool cmp(Rec x, Rec y) {
return x.w < y.w;
}
inline void add(int u, int v, int w) {
t[++cnt].nxt = head[u], t[cnt].ver = v, t[cnt].val = w, head[u] = cnt;
}
inline void Init() {
for (register int i = 1; i <= n; i++)
fa[i] = i;
}
inline void kruskal() {
sort(e + 1, e + m + 1, cmp);
for (register int i = 1; i <= m; i++) {
int eu = find(e[i].u), ev = find(e[i].v);
if (eu == ev) continue;
add(e[i].u, e[i].v, e[i].w), add(e[i].v, e[i].u, e[i].w);
fa[ev] = eu;
++tot;
if (tot == n - 1) break;
}
}
void DFS(int u, int fa, int w) {
deep[u] = deep[fa] + 1;
f[u][0] = fa, h[u][0] = w;
for (register int i = 1; i <= 20; i++)
f[u][i] = f[f[u][i - 1]][i - 1],
h[u][i] = max(h[u][i - 1], h[f[u][i - 1]][i - 1]);
for (register int i = head[u]; i; i = t[i].nxt) {
int v = t[i].ver;
if (v != fa)
DFS(v, u, t[i].val);
}
}
inline int LCA(int u, int v) {
int ans = 0;
if (deep[u] < deep[v]) swap(u, v);
for (register int i = 20; i >= 0; i--)
if (deep[f[u][i]] >= deep[v]) {
ans = max(ans, h[u][i]);
u = f[u][i];
}
if (u == v) return ans;
for (register int i = 20; i >= 0; i--)
if (f[u][i] != f[v][i]) {
ans = max(ans, h[u][i]), ans = max(ans, h[v][i]);
u = f[u][i], v = f[v][i];
}
ans = max(ans, h[u][0]), ans = max(ans, h[v][0]);
return ans;
}
int main() {
//freopen("a.in", "r", stdin);
//freopen("1.out", "w", stdout);
n = read(), m = read();
for (register int i = 1; i <= m; i++)
e[i].u = read(), e[i].v = read(), e[i].w = read();
Init();
kruskal();
for (register int i = 1; i <= n; i++)
if (!deep[i])
DFS(i, 0, 0);
q = read();
for (register int i = 1; i <= q; i++) {
u = read(), v = read();
if (find(u) != find(v))
printf("impossible\n");
else
printf("%d\n", LCA(u, v));
}
return 0;
}
标签:int,max,Number,deep,read,LuoGu8482,ans,getchar
From: https://www.cnblogs.com/Agakiss/p/16623850.html