BZOJ 2303 方格染色(带权并查集)
要使得每个2*2的矩形有奇数个红色,如果我们把红色记为1,蓝色记为0,那么我们得到了这2*2的矩形里的数字异或和为1.
对于每个方格则有a(i,j)^a(i-1,j)^a(i,j-1)^a(i-1,j-1)=1.由这些方程可以推出对于每个方格:
如果i,j都是偶数,则有a(i,j)^a(1,1)^a(i,1)^a(1,j)=1.
否则,a(i,j)^a(1,1)^a(i,1)^a(1,j)=0.枚举a(1,1)的染色情况。可以由a(i,j)的染色情况推出a(i,1)和a(1,j)是否颜色相同或者相反。
类似于a bug's life。那么把它们用带权并查集维护后,染色的种数就是一些联通块的染色种数,因为这些联通块互相不影响。
那么总染色数就是2^(cnt-1).
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000000 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=200005; //Code begin... struct Node{int l, r, c;}node[N]; int fa[N], dis[N]; LL pow_mod(LL a, LL n, LL mod){ LL ret=1, temp=a%mod; while (n) { if (n&1) ret=ret*temp%mod; temp=temp*temp%mod; n>>=1; } return ret; } int find(int x){ int tmp; if (fa[x]!=x) tmp=find(fa[x]), dis[x]=dis[x]^dis[fa[x]], fa[x]=tmp; return fa[x]; } bool union_set(int x, int y, int d){ int u=find(x), v=find(y); if (u!=v) {dis[u]=dis[y]^d^dis[x]; fa[u]=v; return true;} else return dis[x]^dis[y]^d==0; } int main () { int n, m, k, u, v, flag; LL ans=0; scanf("%d%d%d",&n,&m,&k); FOR(i,1,k) scanf("%d%d%d",&node[i].l,&node[i].r,&node[i].c); mem(dis,0); flag=true; FOR(i,1,n+m-1) fa[i]=i; FOR(i,1,k) { if (node[i].l==1) u=1; else u=m+node[i].l-1; v=node[i].r; if (node[i].l%2==0&&node[i].r%2==0) { if (!union_set(u,v,node[i].c^1)) {flag=false; break;} } else { if (!union_set(u,v,node[i].c)) {flag=false; break;} } } if (flag) { int cnt=0; FOR(i,1,n+m-1) if (fa[i]==i) ++cnt; ans=(ans+pow_mod(2,cnt-1,MOD))%MOD; } mem(dis,0); flag=true; FOR(i,1,n+m-1) fa[i]=i; FOR(i,1,k) { if (node[i].l==1) u=1; else u=m+node[i].l-1; v=node[i].r; if (node[i].l%2==0&&node[i].r%2==0) { if (!union_set(u,v,node[i].c)) {flag=false; break;} } else { if (!union_set(u,v,node[i].c^1)) {flag=false; break;} } } if (flag) { int cnt=0; FOR(i,1,n+m-1) if (fa[i]==i) ++cnt; ans=(ans+pow_mod(2,cnt-1,MOD))%MOD; } printf("%lld\n",ans); return 0; }