| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12601 | Accepted: 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:
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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
Source
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;
}原文：https://www.cnblogs.com/autoint/p/10960274.html