BZOJ4668: 冷战 (并查集 + LCA)

题意:动态给点连边 询问两个点之间最早是在第几个操作连起来的

题解:因为并查集按秩合并 秩最高是logn的 所以我们可以考虑把秩看作深度 跑LCA

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 5; int n, m, cnt;
int fa[MAXN];
int id[MAXN];
int zhi[MAXN]; int find(int x) {
if(fa[x]) return find(fa[x]);
else return x;
} int ffind(int x) {
if(fa[x]) return ffind(fa[x]) + 1;
else return 0;
} void add(int x, int y, int z) {
int fx = find(x);
int fy = find(y);
if(fx != fy) {
if(zhi[fx] < zhi[fy]) {
fa[fx] = fy; id[fx] = z;
} else fa[fy] = fx, id[fy] = z;
if(zhi[fx] == zhi[fy]) zhi[fx]++;
}
} int query(int x, int y) {
int fx = find(x);
int fy = find(y);
if(fx != fy) return 0;
int dx = ffind(x), dy = ffind(y);
int res = 0;
while(dx > dy) res = max(res, id[x]), x = fa[x], dx--;
while(dx < dy) res = max(res, id[y]), y = fa[y], dy--;
while(x != y) {
res = max(res, max(id[x], id[y]));
x = fa[x]; y = fa[y];
}
return res;
} int main() {
cnt = 0;
scanf("%d%d", &n, &m);
int las = 0;
for(int i = 1; i <= m; i++) {
int opt, u, v;
scanf("%d%d%d", &opt, &u, &v);
u ^= las, v ^= las;
if(opt) printf("%d\n", las = query(u, v));
else add(u, v, ++cnt);
}
return 0;
}
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