Codeforces Gym 100814C Connecting Graph (并查集, 树链剖分)

题目大意：

就是现在初始的时候有一个只有n个点的图(n <= 1e5), 现在进行m( m <= 1e5 )次操作

每次操作要么添加一条无向边, 要么询问之前结点u和v最早在哪一次操作的时候连通了

大致思路：

这个题表示只想到了O(m*logn*logn)的做法....

首先用并查集维护连通性

当添加边的时候，如果两个点在不同连通块，就选取其连通块的代表结点连一条边, 权值为该操作的编号, 否则不作操作.

那么在n次操作之后, 会得到一个森林

然后我们离线处理询问

如果操作为询问, 则就相当于问树上两个点路径上的最大边权值就是答案(同一连通块, 也就是同一棵树的情况)不在一棵树上就是-1

如果操作为添加边, 那么就把之前的添加的边的权值设置为-1即可

对于边上边权的修改和树的路径的最大边权询问, 树链剖分即可解决

貌似有更简单的方法....不过表示只想到了这个比较明显的做法....并查集只会用来表示连通性好捉急....

代码如下：

Result  :  Accepted     Memory  :  13412 KB     Time  :  638 ms

/* * Author: Gatevin * Created Time:  2015/11/21 14:02:38 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 100010int top[maxn];int grandson[maxn];int dep[maxn];int siz[maxn];int belong[maxn];int father[maxn];int Q[maxn];int cnt;int hson[maxn];int q, s;bool vis[maxn];int T, n, m;int id[maxn];int antiID[maxn];struct Edge{    int u, v, w, nex;    Edge(int _u, int _v, int _w, int _nex)    {        u = _u, v = _v, w = _w, nex = _nex;    }    Edge(){}};int head[maxn];int tot;Edge edge[maxn << 1];int w[maxn];int idx;void add_Edge(int x, int y, int w){    edge[++tot] = Edge(x, y, w, head[x]);    head[x] = tot;}void split(int root){    int l = 0, r = 1;    dep[Q[r] = root] = 1;    father[root] = -1;    w[root] = 0;    while(l < r)    {        int x = Q[++l];        if(head[x] == -1) continue;        for(int j = head[x]; j + 1; j = edge[j].nex)        {            int y = edge[j].v;            if(y == father[x]) continue;            w[y] = edge[j].w;            dep[Q[++r] = y] = dep[x] + 1;            father[y] = x;        }    }    for(int i = r ; i ; i--)    {        int x = Q[i], p = -1;        siz[x] = 1;        if(head[x] == -1) continue;        for(int j = head[x]; j + 1; j = edge[j].nex)        {            int y = edge[j].v;            if(y == father[x]) continue;            siz[x] += siz[y];            if(p == -1 || (p > 0 && siz[y] > siz[p]))                p = y;        }        if(p == -1)        {            hson[x] = -1;            grandson[++cnt] = x;            belong[top[cnt] = x] = cnt;        }        else        {            hson[x] = p;            belong[x] = belong[p];            top[belong[x]] = x;        }    }    //int idx = 0;    //memset(vis, 0, sizeof(vis));    for(int i = r; i; i--)    {        int x = Q[i];        if(vis[x]) continue;        vis[x] = 1;        id[x] = ++idx;        antiID[idx] = x;        while(father[x] != -1 && belong[father[x]] == belong[x] && !vis[father[x]])        {            x = father[x];            id[x] = ++idx;            antiID[idx] = x;            vis[x] = 1;        }    }    return;}#define lson l, mid, rt << 1#define rson mid + 1, r , rt << 1 | 1int val[maxn << 2];void pushUp(int rt){    val[rt] = max(val[rt << 1], val[rt << 1 | 1]);    return;}void build(int l, int r, int rt){    if(l == r)    {        val[rt] = w[antiID[l]];        return;    }    int mid = (l + r) >> 1;    build(lson);    build(rson);    pushUp(rt);}void update(int l, int r, int rt, int pos, int value){    if(l == r)    {        val[rt] = value;        return;    }    int mid = (l + r) >> 1;    if(pos <= mid) update(lson, pos, value);    else update(rson, pos, value);    pushUp(rt);    return;}int query(int l, int r, int rt, int L, int R){    if(l >= L && r <= R)        return val[rt];    int mid = (l + r) >> 1;    int ret = 0;    if(mid >= L) ret = max(ret, query(lson, L, R));    if(mid + 1 <= R) ret = max(ret, query(rson, L, R));    return ret;}int answer(int x, int y){    int ans= 0;    while(top[belong[x]] != top[belong[y]])    {        if(dep[top[belong[x]]] < dep[top[belong[y]]])            swap(x, y);        ans = max(ans, query(1, n, 1, id[x], id[top[belong[x]]]));        x = father[top[belong[x]]];    }    if(x == y) return ans;    if(dep[x] < dep[y]) swap(x, y);    ans = max(ans, query(1, n, 1, id[x], id[hson[y]]));    return ans;}void change(int x, int w){    x <<= 1;    int u = edge[x].u, v = edge[x].v;    if(father[u] == v)        update(1, n, 1, id[u], w);    else update(1, n, 1, id[v], w);    return;}struct Op{    int type, u, v, fu, fv, edge;    bool add;    Op(int _t, int _u, int _v)    {        type = _t, u = _u, v = _v;        add = false;    }    Op(){}};Op op[maxn];int fa[maxn];int find(int x){    return x == fa[x] ? x : fa[x] = find(fa[x]);}int E;//添加的边的次数void Union(int x, int y, int opid){    int fx = find(x);    int fy = find(y);    op[opid].fu = fx, op[opid].fv = fy;    op[opid].add = false;    if(fx != fy)    {        fa[fx] = fy;        op[opid].add = true;        op[opid].edge = ++E;        add_Edge(fx, fy, opid);        add_Edge(fy, fx, opid);    }}int ans[maxn];int read(){    int x = 0;    char c;    while(!isdigit(c = getchar())) continue;    x = (x << 3) + (x << 1) + c - '0';    while(isdigit(c = getchar())) x = (x << 3) + (x << 1) + c - '0';    return x;}int sis[maxn];int main(){    //scanf("%d", &T);    T = read();    int cas = 0;    while(T--)    {        cas++;        //scanf("%d %d", &n, &m);        n = read(), m = read();        memset(head, -1, sizeof(head));        tot = 0; E = 0;        for(int i = 1; i <= n; i++) fa[i] = i;        for(int i = 1; i <= m; i++)        {            //scanf("%d %d %d", &op[i].type, &op[i].u, &op[i].v);            op[i].type = read(), op[i].u = read(), op[i].v = read();            if(op[i].type == 1) Union(op[i].u, op[i].v, i);        }        idx = 0;        memset(vis, 0, sizeof(vis));        cnt = 0;        for(int i = 1; i <= n; i++)        {            int fi = find(i);            if(sis[fi] != cas)            {                split(fi);                sis[fi] = cas;            }        }        build(1, n, 1);        for(int i = m; i > 0; i--)        {            if(op[i].type == 2)//query            {                int u = op[i].u, v = op[i].v;                if(find(u) != find(v))//一直就是不连通的                {                    ans[i] = -1;                    continue;                }                ans[i] = answer(op[i].u, op[i].v);            }            else            {                if(op[i].add == false) continue;//这个操作没有造成过添加边的影响                change(op[i].edge, 1e9);            }        }        for(int i = 1; i <= m; i++)            if(op[i].type == 2)                printf("%d\n", ans[i] == 1e9 ? -1 : ans[i]);    }    return 0;}