ZOJ 3656 Bit Magic (2-Sat問題)

題意：b[i][j] 與a[i], a[j]的關係如下。現在已知b，問是否存在對應的a，若存在輸出YES，否則輸出NO

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];}}}

題解：

#include<cstring>#include<algorithm>#include <cstdio>using namespace std;#define MAXN 1010int n;unsigned int b[510][510];int m[1010][1010];int id[1010];int find_components(int n,int mat[][MAXN],int* id){int ret=0,a[MAXN],b[MAXN],c[MAXN],d[MAXN],i,j,k,t;for (k=0;k<n;id[k++]=0);for (k=0;k<n;k++)if (!id[k]){for (i=0;i<n;i++)a[i]=b[i]=c[i]=d[i]=0;a[k]=b[k]=1;for (t=1;t;)for (t=i=0;i<n;i++){if (a[i]&&!c[i])for (c[i]=t=1,j=0;j<n;j++)if (mat[i][j]&&!a[j])a[j]=1;if (b[i]&&!d[i])for (d[i]=t=1,j=0;j<n;j++)if (mat[j][i]&&!b[j])b[j]=1;}for (ret++,i=0;i<n;i++)if (a[i]&b[i])id[i]=ret;}return ret;}void build_map(int k){    memset(m, 0, sizeof(m));    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++)        {            if(i == j) continue;            if(i % 2 == 1 && j % 2 == 1)            {                if(b[i][j] & (1<<k))                {                    m[i][j+n] = 1;                    m[j][i+n] = 1;                }                else                {                    m[i+n][i] = 1;                    m[j+n][j] = 1;                }            }            else if(i % 2 == 0 && j % 2 == 0)            {                if(b[i][j] & (1<<k))                {                    m[i][i+n] = 1;                    m[j][j+n] = 1;                }                else                {                    m[j+n][i] = 1;                    m[i+n][j] = 1;                }            }            else            {                if(b[i][j] & (1<<k) )                {                    m[i][j+n] = m[j+n][i] = 1;                    m[j][i+n] = m[i+n][j] = 1;                }                else                {                    m[i][j] = m[j][i] = 1;                    m[i+n][j+n] = m[j+n][i+n] = 1;                }            }        }}void solve(){    for(int k = 0; k < 32; k++)    {        build_map(k);        memset(id, 0, sizeof(id));        find_components(2*n, m, id);        for(int i = 0; i < n; i++)            if(id[i] == id[i+n])            {                printf("NO\n");                return;            }    }    printf("YES\n");}int main(){while(scanf("%d",&n) != EOF)    {    int i, j;    memset(id, 0, sizeof(id));    memset(b, 0, sizeof(b));        for(i = 0; i < n; i++)            for(j = 0; j < n; j++)                scanf("%d",&b[i][j]);        bool flag = true;        for(i = 0; i < n; i++)            if(b[i][i]) { flag = false; break; }        if(flag == false)        {            printf("NO\n");            continue;        }        solve();    }return 0;}