POJ 3678 Katu Puzzle

                                                                     Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11429		Accepted: 4233

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 V_i a value X_i (0 ≤ X_i ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = c

The calculating rules are:

AND 0 1 0 0 0 1 0 1

OR 0 1 0 0 1 1 1 1

XOR 0 1 0 0 1 1 1 0

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

X ₀ = 1,  X ₁ = 1,  X ₂ = 0,  X ₃ = 1.

 

建图有问题，其他的很简单，容我仔细思考一下。

 

#include<iostream>
#include<vector>
#include<stack>
#include<cstdio>
using namespace std;
int n,m;
vector<int>u[200024];
stack<int>st;
int dfn[200024],sig,low[200024],color[20024],index;
bool book[200024];
void init()
{
    scanf("%d%d",&n,&m);
    char s[10];
    int x,y,c;
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&x,&y,&c);
        scanf("%s",s);
        if(s[0]=='A'){
            if(c==1){
                u[y].push_back(y+n);//不理解
                u[x].push_back(x+n);//不理解
            }
            if(c==0){
                u[x+n].push_back(y);
                u[y+n].push_back(x);
            }
        }
        else if(s[0]=='X'){
            if(c==1){
                u[x+n].push_back(y);
                u[x].push_back(y+n);
                u[y+n].push_back(x);
                u[y].push_back(x+n);
            }
            if(c==0){
                u[x+n].push_back(y+n);
                u[x].push_back(y);
                u[y+n].push_back(x+n);
                u[y].push_back(x);
            }
        }
        else if(s[0]=='O'){
            if(c==1){
                u[x].push_back(y+n);
                u[y].push_back(x+n);
            }
            if(c==0){
                u[y+n].push_back(y);//不理解
                u[x+n].push_back(x);//不理解
            }
        }

    }
}

void tarjan(int t)
{
    dfn[t]=low[t]=++index;
    book[t]=true;
    st.push(t);
    int siz=u[t].size();
    for(int i=0;i<siz;i++){
        if(!dfn[u[t][i]]){
            tarjan(u[t][i]);
            low[t]=min(low[t],low[u[t][i]]);
        }
        if(book[u[t][i]]){
            low[t]=min(low[t],low[u[t][i]]);
        }
    }
    int miui;
    if(dfn[t]==low[t]){
        sig++;
        while(true){
            if(st.empty()){break;}
            miui=st.top();
            st.pop();
            book[miui]=false;
            color[miui]=sig;
            if(miui==t){break;}
        }
    }
}


bool solve()
{
    for(int i=0;i<n;i++){
        if(!dfn[i]){
            tarjan(i);
        }
    }

    for(int i=0;i<n;i++){
        if(color[i]==color[i+n]&&color[i]!=0){
            return false;
        }
    }
    return true;
}

int main()
{
    init();
    if(solve()){
        printf("YES\n");
    }
    else printf("NO\n");
}