POJ 3678 Katu Puzzle
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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 "); } else printf("NO "); }
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