Yesterday, my teacher taught me about bit operators: and (&), or (|), xor (^). I generated a number table a[N], and wrote a program to calculate the matrix table b[N][N] using three kinds of bit operator. I thought my achievement would get teacher's attention.
The key function is the code showed below.
void calculate(int a[N], int b[N][N]) {
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
if (i == j) b[i][j] = 0;
else if (i % 2 == 1 && j % 2 == 1) b[i][j] = a[i] | a[j];
else if (i % 2 == 0 && j % 2 == 0) b[i][j] = a[i] & a[j];
else b[i][j] = a[i] ^ a[j];
}
}
}
There is no doubt that my teacher raised lots of interests in my work and was surprised to my talented programming skills. After deeply thinking, he came up with another problem: if we have the matrix table b[N][N] at first, can you check whether corresponding number table a[N] exists?
Input
There are multiple test cases.
For each test case, the first line contains an integer N, indicating the size of the matrix. (1 ≤ N ≤ 500).
The next N lines, each line contains N integers, the jth integer in ith line indicating the element b[i][j] of matrix. (0 ≤ b[i][j] ≤ 2 ^ 31 - 1)
Output
For each test case, output "YES" if corresponding number table a[N] exists; otherwise output "NO".
Sample Input
2 0 4 4 0 3 0 1 24 1 0 86 24 86 0
Sample Output
YES NO
本题很简单，明显的2-SAT的模版题。
长春现场赛时一眼就看出了这题是2-SAT。只不过模板写错了一点，多加了个分号，导致样例一直出不来。幸好调试之后发现，
修改一下，交上去1A，爽~~~~此题是做得最顺利的了。
现场赛的时候内存真的是无穷大啊，直接做31*500*2=31000个点的2-SAT就AC了。但是比赛结束后在ZOJ上就会MLE了~~~~~
要分开做，做31次的2-SAT就可以了~~~~~~
幸好现场赛没有发现这个问题~~~给力啊~！！！
#include<stdio.h> #include<iostream> #include<algorithm> #include<vector> #include<queue> #include<string.h> using namespace std; const int MAXN=1100;// bool visit[MAXN]; queue<int>q1,q2; //vector建图方法很妙 vector<vector<int> >adj; //原图 //中间一定要加空格把两个'>'隔开 vector<vector<int> >radj;//逆图 vector<vector<int> >dag;//缩点后的逆向DAG图 int n,m,cnt; int id[MAXN],order[MAXN],ind[MAXN];//强连通分量，访问顺序，入度 void dfs(int u) { visit[u]=true; int i,len=adj[u].size(); for(i=0;i<len;i++) if(!visit[adj[u][i]]) dfs(adj[u][i]); order[cnt++]=u; } void rdfs(int u) { visit[u]=true; id[u]=cnt; int i,len=radj[u].size(); for(i=0;i<len;i++) if(!visit[radj[u][i]]) rdfs(radj[u][i]); } void korasaju() { int i; memset(visit,false,sizeof(visit)); for(cnt=0,i=0;i<2*n;i++) if(!visit[i]) dfs(i); memset(id,0,sizeof(id)); memset(visit,false,sizeof(visit)); for(cnt=0,i=2*n-1;i>=0;i--) if(!visit[order[i]]) { cnt++;//这个一定要放前面来 rdfs(order[i]); } } bool solvable() { for(int i=0;i<n;i++) if(id[2*i]==id[2*i+1]) return false; return true; } void add(int x,int y) { adj[x].push_back(y); radj[y].push_back(x); } int b[600][600]; int bit[33]; int main() { int N; bit[0]=1; for(int i=1;i<31;i++)bit[i]=2*bit[i-1]; while(scanf("%d",&N)!=EOF) { n=N; bool flag=true; for(int i=0;i<N;i++) for(int j=0;j<N;j++) { scanf("%d",&b[i][j]); if(i==j&&b[i][j]!=0)flag=false; } if(!flag) { printf("NO\n"); continue; } for(int i=0;i<N;i++) { if(!flag)break; for(int j=i+1;j<N;j++) if(b[i][j]!=b[j][i]) { flag=false; break; } } if(!flag) { printf("NO\n"); continue; } for(int k=0;k<31;k++) { adj.assign(2*n,vector<int>()); radj.assign(2*n,vector<int>()); for(int i=0;i<N;i++) for(int j=i+1;j<N;j++) { if(i%2==1&&j%2==1) { int t1=i; int t2=j; if(b[i][j]&bit[k]) { add(2*t1,2*t2+1); add(2*t2,2*t1+1); } else { add(2*t1+1,2*t1); add(2*t2+1,2*t2); } } else if(i%2==0&&j%2==0) { int t1=i; int t2=j; if(b[i][j]&bit[k]) { add(2*t1,2*t1+1); add(2*t2,2*t2+1); } else { add(2*t1+1,2*t2); add(2*t2+1,2*t1); } } else { int t1=i; int t2=j; if(b[i][j]&bit[k]) { add(2*t1,2*t2+1); add(2*t1+1,2*t2); add(2*t2,2*t1+1); add(2*t2+1,2*t1); } else { add(2*t1,2*t2); add(2*t1+1,2*t2+1); add(2*t2,2*t1); add(2*t2+1,2*t1+1); } } } korasaju(); if(!solvable()) { flag=false; break; } } if(flag)printf("YES\n"); else printf("NO\n"); } return 0; }