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  • 图论(2-sat):HDU 4421 Bit Magic

    Bit Magic

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3040    Accepted Submission(s): 871


    Problem Description
    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.

    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
      注意数组要开大。
      貌似并查集也可以做。
      1 #include <iostream>
      2 #include <cstring>
      3 #include <cstdio>
      4 using namespace std;
      5 const int maxn=1010;
      6 const int maxm=800010;
      7 int n,cnt,fir[maxn],to[maxm*2],nxt[maxm*2];
      8 int tot,scnt,ID[maxn],low[maxn],scc[maxn];
      9 
     10 int min(int a,int b){
     11     return a<b?a:b;
     12 }
     13 
     14 void addedge(int a,int b){
     15     nxt[++cnt]=fir[a];
     16     fir[a]=cnt;
     17     to[cnt]=b;
     18 }
     19 
     20 bool Inst[maxn];
     21 int st[maxn],top;
     22 void Tarjan(int x){
     23     low[x]=ID[x]=++tot;
     24     st[++top]=x;Inst[x]=true;
     25     for(int i=fir[x];i;i=nxt[i])
     26         if(!ID[to[i]]){
     27             Tarjan(to[i]);
     28             low[x]=min(low[x],low[to[i]]);
     29         }
     30         else if(Inst[to[i]])
     31             low[x]=min(low[x],ID[to[i]]);
     32     if(low[x]==ID[x]){
     33         ++scnt;
     34         while(true){
     35             int y=st[top--];
     36             scc[y]=scnt;
     37             Inst[y]=false;
     38             if(x==y)break;
     39         }        
     40     }
     41 }
     42 
     43 bool Check(){
     44     for(int i=0;i<n*2;i++)
     45         if(!ID[i])Tarjan(i);
     46     for(int i=0;i<n;i++)
     47         if(scc[i*2]==scc[i*2+1])
     48             return false;
     49     return true;            
     50 }
     51 
     52 int b[510][510];
     53 void Init(){
     54     memset(fir,0,sizeof(fir));
     55     memset(scc,0,sizeof(scc));
     56     memset(ID,0,sizeof(ID));
     57     cnt=tot=scnt=0;
     58 }
     59 
     60 bool Judge(){
     61     for(int t=0;t<=29;t++){
     62         Init();
     63         for(int i=0;i<n;i++){
     64             for(int j=0;j<n;j++){
     65                 if(i==j){if(b[i][j])return false;continue;}
     66                 else if((i&1)&&(j&1)){
     67                     if(b[i][j]&(1<<t)){
     68                         addedge(i*2,j*2+1);
     69                         addedge(j*2,i*2+1);
     70                     }
     71                     else{
     72                         addedge(i*2+1,i*2);//这两句并不影响答案,有谁知道为啥
     73                         addedge(j*2+1,j*2);//
     74                         addedge(i*2,j*2);
     75                         addedge(j*2,i*2);
     76                     }
     77                 }
     78                 else if(!((i&1)||(j&1))){
     79                     if(b[i][j]&(1<<t)){
     80                         addedge(i*2,i*2+1);
     81                         addedge(j*2,j*2+1);
     82                         addedge(i*2+1,j*2+1);
     83                         addedge(j*2+1,i*2+1);                    
     84                     }
     85                     else{
     86                         addedge(i*2+1,j*2);
     87                         addedge(j*2+1,i*2);
     88                     }
     89                 }
     90                 else{
     91                     if(b[i][j]&(1<<t)){
     92                         addedge(i*2+1,j*2);
     93                         addedge(j*2+1,i*2);
     94                         addedge(i*2,j*2+1);
     95                         addedge(j*2,i*2+1);
     96                     }
     97                     else{
     98                         addedge(i*2,j*2);
     99                         addedge(j*2,i*2);
    100                         addedge(i*2+1,j*2+1);
    101                         addedge(j*2+1,i*2+1);
    102                     }
    103                 }
    104             }    
    105         }
    106         if(!Check())
    107             return false;
    108     }
    109     return true;
    110 }
    111 
    112 int main(){
    113     while(scanf("%d",&n)!=EOF){
    114         for(int i=0;i<n;i++)
    115             for(int j=0;j<n;j++)
    116                 scanf("%d",&b[i][j]);
    117         if(Judge())
    118             printf("YES
    ");
    119         else
    120             printf("NO
    ");    
    121     }
    122     return 0;
    123 }
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  • 原文地址:https://www.cnblogs.com/TenderRun/p/5634173.html
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