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  • POJ3678 Katu Puzzle

    Katu Puzzle

    Katu Puzzle
    Time Limit: 1000MSMemory Limit: 65536K
    Total Submissions: 12601Accepted: 4627

    Description

    Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

     Xa op Xb = c

    The calculating rules are:

    AND01
    000
    101
    OR01
    001
    111
    XOR01
    001
    110

    Given a Katu Puzzle, your task is to determine whether it is solvable.

    Input

    The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
    The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

    Output

    Output a line containing "YES" or "NO".

    Sample Input

    4 4
    0 1 1 AND
    1 2 1 OR
    3 2 0 AND
    3 0 0 XOR

    Sample Output

    YES

    Hint

    X0 = 1, X1 = 1, X2 = 0, X3 = 1.

    Source

    题解

    2-SAT经典题,连边如下:

    A AND B = 1   !A->A   !B->B
    A AND B = 0   A->!B   B->!A
    A OR B = 1      !A->B   !B->A
    A OR B = 0      A->!A   B->!B
    A XOR B = 1   A->!B !B->A !A->B B->!A
    A XOR B = 0   A->B B->A !A->!B !B->!A

    然后tarjan求SCC判断有无解即可。

    时间复杂度(O(n+m))

    #include<iostream>
    #include<vector>
    #define rg register
    #define il inline
    #define co const
    template<class T>il T read(){
        rg T data=0,w=1;rg char ch=getchar();
        for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
        for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
        return data*w;
    }
    template<class T>il T read(rg T&x) {return x=read<T>();}
    typedef long long ll;
    using namespace std;
    
    co int N=1006;
    int n,m,dfn[N],low[N],num,st[N],top,c[N],cnt;
    bool ins[N];
    vector<int> e[N*2];
    
    il void add(int x,int y){
    	e[x].push_back(y);
    }
    void tarjan(int x){
    	dfn[x]=low[x]=++num;
    	st[++top]=x,ins[x]=1;
    	for(unsigned i=0;i<e[x].size();++i){
    		int y=e[x][i];
    		if(!dfn[y]){
    			tarjan(y);
    			low[x]=min(low[x],low[y]);
    		}
    		else if(ins[y]) low[x]=min(low[x],dfn[y]);
    	}
    	if(dfn[x]==low[x]){
    		++cnt;
    		int y;
    		do y=st[top--],ins[y]=0,c[y]=cnt;
    		while(x!=y);
    	}
    }
    int main(){
    	read(n),read(m);
    	for(int a,b,c;m--;){
    		char s[6];
    		read(a),read(b),read(c),scanf("%s",s);
    		if(s[0]=='A'){
    			if(c) add(a,a+n),add(b,b+n);
    			else add(a+n,b),add(b+n,a);
    		}
    		else if(s[0]=='O'){
    			if(c) add(a,b+n),add(b,a+n);
    			else add(a+n,a),add(b+n,b);
    		}
    		else{
    			if(c) add(a,b+n),add(b,a+n),add(a+n,b),add(b+n,a);
    			else add(a,b),add(b,a),add(a+n,b+n),add(b+n,a+n);
    		}
    	}
    	for(int i=0;i<n;++i)
    		if(!dfn[i]) tarjan(i);
    	for(int i=0;i<n;++i)
    		if(c[i]==c[i+n]) return puts("NO"),0;
    	return puts("YES"),0;
    }
    
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  • 原文地址:https://www.cnblogs.com/autoint/p/10960274.html
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