BZOJ4668——冷战

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1、题意：给一些点，有两种操作，连通两个点和询问两个点最早是在什么时候连通
2、分析：这题用lct维护动态最小生成树就好。。但是过于暴力，我们可以倒着用并查集暴力查询，然后依旧是查询链上的最大值，然而这次我们不路径压缩，我们按秩合并，然后树高logn，接下来就随意弄啦

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define M 1000010inline int read(){    char ch = getchar(); int x = 0, f = 1;    while(ch < '0' || ch > '9'){        if(ch == '-') f = -1;        ch = getchar();    }    while('0' <= ch && ch <= '9'){        x = x * 10 + ch - '0';        ch = getchar();    }    return x * f;}int fa[M], Rank[M], dep[M], val[M]; inline int find(int x){    if(fa[x] == x) return x;    int ret = find(fa[x]);    dep[x] = dep[fa[x]] + 1;    return ret;}inline int lca(int x, int y){    int ans = 0;    while(x != y){        if(dep[x] < dep[y]) swap(x, y);        ans = max(ans, val[x]);        x = fa[x];    }    return ans;}int main(){    //freopen("0input.in", "r", stdin);     int n = read(), m = read();    for(int i = 1; i <= n; i ++){        fa[i] = i; Rank[i] = 1; dep[i] = 0;    }    int time = 0, ans = 0;    for(int i = 1; i <= m; i ++){        int op = read(), x = read(), y = read();        x ^= ans; y ^= ans;        if(!op){            int px = find(x), py = find(y);            time ++;            if(px != py){                if(Rank[px] >= Rank[py]){                    fa[py] = px;                    val[py] = time;                    if(Rank[px] == Rank[py]) Rank[px] ++;                }                else{                    fa[px] = py;                    val[px] = time;                }            }        }        else{            int px = find(x), py = find(y);            if(px != py) ans = 0;            else ans = lca(x, y);            printf("%d\n", ans);        }    }    return 0;}

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