bnuoj 51275 并查集按深度合并建树

道路修建 Large

Time Limit: 5000ms

Memory Limit: 131072KB

64-bit integer IO format: %lld      Java class name: Main

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无向图初始有个点，从到依次标号，但是没有边，

接下来有次操作，从到依次标号，你需要对每种操作输出相应的结果，操作分为两种：

输入格式	操作说明	输出结果
0_u_v	加入一条连接标号为和标号为的点的边。	输出加边后图中连通块的个数。
1_u_v	查询标号为和标号为的点之间是否连通。	如果连通，输出，表示最早在第次操作后标号为和标号为的点之间连通，否则输出。

（输入格式中的下划线‘_’表示实际输入文件中的空格）

 

Input

第一行是一个正整数，表示测试数据的组数，

对于每组测试数据，

第一行包含两个整数、，

接下来行，每行是个整数、、，请注意所给的、均是经过加密的，

解密方式是、 ，其中表示上一次操作的输出结果，

初始，保证，解密后且。

 

Output

对于每组测试数据，

输出行，每行包含一个整数，表示操作的输出结果。

 

Sample Input

1
4 7
0 1 2
1 1 0
0 1 3
0 0 1
1 0 1
0 1 7
1 0 5

Sample Output

3
0
2
2
3
1
6

Source

quailty's Contest #1

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;

int n,m;
int fa[maxn],val[maxn];
int rk[maxn];
int op,u,v;
int vis[maxn];

int find(int x)
{
    return fa[x]==x?x:find(fa[x]);
}

int main()
{
    freopen("in.txt","r",stdin);
    int T;cin>>T;
    while(T--){
        scanf("%d%d",&n,&m);
        REP(i,1,n) fa[i]=i,val[i]=0;
        REP(i,1,n) rk[i]=1;
        REP(i,1,n) vis[i]=0;
        int lans=0,cnt=n;
        REP(i,1,m){
            scanf("%d%d%d",&op,&u,&v);
            u^=lans;v^=lans;
            int x=find(u),y=find(v);
            if(op==0){
                if(x!=y){
                    if(rk[x]>rk[y]) swap(x,y);
                    fa[x]=y;
                    rk[y]=max(rk[y],rk[x]+1);
                    val[x]=i;
                    cnt--;
                }
                lans=cnt;
            }
            else{
                if(x!=y) lans=0;
                else{
                    int t=u;
                    vis[x]=i;
                    while(fa[t]!=t) vis[t]=i,t=fa[t];
                    int lca=x;
                    t=v;
                    while(fa[t]!=t){
                        if(vis[t]==i){
                            lca=t;break;
                        }
                        t=fa[t];
                    }
                    lans=0;
                    while(u!=lca) lans=max(lans,val[u]),u=fa[u];
                    while(v!=lca) lans=max(lans,val[v]),v=fa[v];
                }
            }
            printf("%d\n",lans);
        }
    }
    return 0;
}