Exclusive-OR

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2177    Accepted Submission(s): 603

Problem Description

You are not given n non-negative integers
X₀, X₁, ..., X_n-1 less than 2²⁰ , but they do exist, and their values never change.

I‘ll gradually provide you some facts about them, and ask you some questions.

There are two kinds of facts, plus one kind of question:

Input

There will be at most 10 test cases. Each case begins with two integers
n and Q (1 <= n <= 20,000, 2 <= Q <= 40,000). Each of the following lines contains either a fact or a question, formatted as stated above. The
k parameter in the questions will be a positive integer not greater than 15, and the
v parameter in the facts will be a non-negative integer less than 2²⁰. The last case is followed by
n=Q=0, which should not be processed.

Output

For each test case, print the case number on its own line, then the answers, one on each one. If you can‘t deduce the answer for a particular question, from the facts I provide you
before that question, print "I don‘t know.", without quotes. If the
i-th fact (don‘t count questions) cannot be consistent with
all the facts before that, print "The first i facts are conflicting.", then keep silence for everything after that (including facts and questions). Print a blank line after the output of each test case.

Sample Input

2 6
I 0 1 3
Q 1 0
Q 2 1 0
I 0 2
Q 1 1
Q 1 0
3 3
I 0 1 6
I 0 2 2
Q 2 1 2
2 4
I 0 1 7
Q 2 0 1
I 0 1 8
Q 2 0 1
0 0

Sample Output

Case 1:
I don‘t know.
3
1
2
Case 2:
4
Case 3:
7
The first 2 facts are conflicting.

Source

2009 Asia Wuhan Regional Contest Hosted by Wuhan University

题意：

有n(n<=20000)个未知的整数X0,X1,X2Xn-1，有以下Q个(Q<=40000)操作：

I p v :告诉你Xp=v

I p q v :告诉你Xp ^ Xq=v

Q k p1 p2 … pk : 询问 Xp1 Xor Xp2 .. Xor Xpk， k不大于15。

如果当前的I跟之前的有冲突的话，跳出

思路：

并查集，每个节点记录与根节点的异或偏移量，一个集合内如果有一个知道值了的话，这个集合里面都能知道值，（可以标记根是否已经得到值，或者像网上大部分人的做法，虚拟出一个节点n，值为0，将I操作统一）查询时，如果一个未知集合出现了偶数个，那么可以得到其值，u^root^v^root=u^v，如果出现奇数次，那么不能得到。Merge的时候有个小坑，见代码。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#include<sstream>
#define maxn 20005
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot,flag;
int pre[maxn],px[maxn],val[maxn],vis[maxn];
vector<int>root,q[16];
char s[5],buff[100];

int Find(int x)
{
    if(x==pre[x]) return x;
    int r=pre[x];
    pre[x]=Find(pre[x]);
    px[x]=px[r]^px[x];
    return pre[x];
}
bool Merge(int u,int v,int w)
{
    int x=Find(u),y=Find(v);
    if(x!=y)
    {
        if(vis[x]||vis[y])// 如果有集合知道值 必须以知道值集合的根为根
        {
            if(!vis[y]) swap(x,y);
        }
        pre[x]=y;
        px[x]=w^px[u]^px[v];
    }
    else
    {
        if((px[u]^px[v])!=w) return false ;
    }
    return true ;
}
void update()
{
    int u,v,w;
    cnt++;
    gets(buff);
    stringstream si;
    si.clear(); si.str(buff);
    si>>u; si>>v;
    if(si>>w)
    {
        u++,v++;
        if(flag) return ;
        if(!Merge(u,v,w))
        {
            flag=1;
            printf("The first %d facts are conflicting.\n",cnt);
        }
    }
    else
    {
        if(flag) return ;
        u++;
        int r=Find(u);
        if(vis[r])
        {
            if(val[r]!=(v^px[u]))
            {
                flag=1;
                printf("The first %d facts are conflicting.\n",cnt);
            }
        }
        else
        {
            vis[r]=1;
            val[r]=v^px[u];
        }
    }
}
void query()
{
    int i,j,u,v,k,r,flg;
    scanf("%d",&k);
    root.clear();
    for(i=0;i<k;i++) q[i].clear();
    for(i=1; i<=k; i++)
    {
        scanf("%d",&u);
        u++;
        int r=Find(u);
        flg=0;
        for(j=0;j<root.size();j++)
        {
            if(r==root[j])
            {
                flg=1;
                q[j].push_back(u);
                break ;
            }
        }
        if(!flg)
        {
            root.push_back(r);
            q[root.size()-1].push_back(u);
        }
    }
    if(flag) return ;
    flg=0;
    int res=0;
    for(i=0;i<root.size();i++)
    {
        if(vis[root[i]])
        {
            for(j=0; j<q[i].size(); j++)
            {
                res^=(px[q[i][j]]^val[root[i]]);
            }
        }
        else
        {
            if(q[i].size()%2==1)
            {
                flg=1;
                break ;
            }
            else
            {
                for(j=0; j<q[i].size(); j+=2)
                {
                    u=px[q[i][j]]^px[q[i][j+1]];
                    res^=u;
                }
            }
        }
    }
    if(flg) printf("I don't know.\n");
    else printf("%d\n",res);
}
int main()
{
    int i,j,t,u,v,w,test=0;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0) break ;
        for(i=1; i<=n; i++)
        {
            pre[i]=i;
            px[i]=vis[i]=0;
        }
        flag=cnt=0;
        printf("Case %d:\n",++test);
        for(i=1; i<=m; i++)
        {
            scanf("%s",s);
            if(s[0]=='I') update();
            else query();
        }
        puts("");
    }
    return 0;
}

hdu 3234 Exclusive-OR （并查集+异或性质）,布布扣,bubuko.com