HDU3234Exclusive-OR(并查集)与HDU3038相似

Exclusive-OR

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2383    Accepted Submission(s): 667

Problem Description

You are not given n non-negative integersX₀, 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 integersn 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. Thek parameter in the questions will be a positive integer not greater than 15, and thev parameter in the facts will be a non-negative integer less than 2²⁰. The last case is followed byn=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 youbefore that question, print "I don't know.", without quotes. If thei-th fact (don't count questions) cannot be consistent withall 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 6I 0 1 3Q 1 0Q 2 1 0I 0 2Q 1 1Q 1 03 3I 0 1 6I 0 2 2Q 2 1 22 4I 0 1 7Q 2 0 1I 0 1 8Q 2 0 10 0

 

Sample Output

Case 1:I don't know.312Case 2:4Case 3:7The first 2 facts are conflicting.

 

Source

2009 Asia Wuhan Regional Contest Hosted by Wuhan University 

注意：逻辑运算的优先级比比较运算符的优先级低。

#include<stdio.h>#include<string.h>const int N = 20005;int fath[N],XOR[N],n,vist[N];int know[N],ans[N];void init(){    for(int i=0;i<=n;i++)    {        fath[i]=i; XOR[i]=0;        know[i]=0; vist[i]=0;    }}int findfath(int x){    if(x!=fath[x])    {        int tx=fath[x];        fath[x]=findfath(fath[x]);        XOR[x]^=XOR[tx];    }    return fath[x];}int set1(int x,int v){    if(know[x])    {        if(ans[x]!=v)return 0;        else return 1;    }    int tx=findfath(x);    if(know[tx])    {       if(XOR[x]!=(ans[tx]^v))            return 0;    }    else        know[tx]=1,ans[tx]=XOR[x]^v;    know[x]=1; ans[x]=v;    return 1;}int set2(int x,int y,int v){    int tx=findfath(x);    int ty=findfath(y);    if(tx==ty)    {        if((XOR[x]^XOR[y])!=v)return 0;        else return 1;    }    if(know[tx])    {        fath[ty]=tx; XOR[ty]=XOR[x]^XOR[y]^v;    }    else    {        fath[tx]=ty; XOR[tx]=XOR[x]^XOR[y]^v;    }    return 1;}int a[N],node[N];int Qoperator(int conflict){    int k,nk=0;    scanf("%d",&k);    for(int i=0;i<k;i++)        scanf("%d",&a[i]);    if(conflict)        return -1;    int flag=0,aa=0;    for(int i=0;i<k;i++)    {        if(know[a[i]])            aa^=ans[a[i]];        else        {            int x=findfath(a[i]);            if(know[x])            {                ans[a[i]]=XOR[a[i]]^ans[x]; know[a[i]]=1;                aa^=ans[a[i]];            }            else            {                node[nk++]=x;                vist[x]++;                aa^=XOR[a[i]];                if(vist[x]%2) flag++;                else flag--;            }        }    }    for(int j=0;j<nk;j++)        vist[node[j]]=0;    if(flag)        return -1;    return aa;}int main(){    int cas=0,m,a,b,v,conflict,flag;    char s[50];    while(scanf("%d%d",&n,&m)>0&&n+m!=0)    {        printf("Case %d:\n",++cas);        init();        conflict=0;        int index=0;        while(m--)        {            scanf("%s",s);            if(s[0]=='I')            {                index++;                scanf("%d%d",&a,&b);                if(getchar()=='\n')                {                    if(conflict)continue;                    flag=set1(a,b);                }                else                {                   scanf("%d",&v);                    if(conflict)continue;                    flag=set2(a,b,v);                }                if(flag==0)                {                    printf("The first %d facts are conflicting.\n",index); conflict=1;                }            }            else            {                flag=Qoperator(conflict);                if(conflict==0)                {                    if(flag==-1)                    printf("I don't know.\n");                    else                    printf("%d\n",flag);                }            }        }        printf("\n");    }}