标签：accept fir input ati cal highlight ons output tarjan
| 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 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
#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");
}
标签：accept fir input ati cal highlight ons output tarjan
原文地址：https://www.cnblogs.com/ZGQblogs/p/9402550.html
