bzoj 4537: [Hnoi2016]最小公倍数 分块

        按照a分成M^0.5块，然后每一块按照b排序。这样查询的时候把所有当前块之前的路径按b排序扫一遍，当前块中的路径暴力插入还原。用并查集维护即可。

       注意常数QAQ。

AC代码如下：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#define N 100005using namespace std;void up(int &x,int y){ if (x<y) x=y; }int read(){int x=0; char ch=getchar();while (ch<'0' || ch>'9') ch=getchar();while (ch>='0' && ch<='9'){ x=x*10+ch-'0'; ch=getchar(); }return x;}struct edg{int x,y,u,v,k;void getin(){x=read(); y=read(); u=read(); v=read();}}a[N],b[N],c[N],h[N];int n,m,cnt,tot,fa[N],sz[N],fu[N],fv[N]; bool ans[N];bool cmpu(const edg &p,const edg &q){ return p.u<q.u || p.u==q.u && p.v<q.v; }bool cmpv(const edg &p,const edg &q){ return p.v<q.v || p.v==q.v && p.u<q.u; }int getfa(int x){ return (x==fa[x])?x:getfa(fa[x]); }void mrg(int x,int y,int u,int v){x=getfa(x); y=getfa(y); if (sz[x]>sz[y]) swap(x,y);h[++tot].x=x; h[tot].y=y; h[tot].k=sz[y];h[tot].u=fu[y]; h[tot].v=fv[y];if (x!=y){fa[x]=y; sz[y]+=sz[x];up(fu[y],fu[x]); up(fv[y],fv[x]);}up(fu[y],u); up(fv[y],v);}int main(){n=read(); m=read(); int i;for (i=1; i<=m; i++) a[i].getin();sort(a+1,a+m+1,cmpu);cnt=read();for (i=1; i<=cnt; i++){ b[i].getin(); b[i].k=i; }sort(b+1,b+cnt+1,cmpv);int blk=(int)sqrt(m),j,k,l,len,p,q;for (i=1; i<=m; i+=blk){for (j=1; j<=n; j++){ fa[j]=j; fu[j]=fv[j]=-1; sz[j]=1; }len=0;for (j=1; j<=cnt; j++)if (b[j].u>=a[i].u && (i+blk>m || b[j].u<a[i+blk].u)) c[++len]=b[j];if (!len) continue;if (i>1) sort(a+1,a+i,cmpv);for (j=k=1; j<=len; j++){for (; k<i && a[k].v<=c[j].v; k++)mrg(a[k].x,a[k].y,a[k].u,a[k].v);tot=0;for (l=i; l<i+blk && l<=m; l++)if (a[l].u<=c[j].u && a[l].v<=c[j].v)mrg(a[l].x,a[l].y,a[l].u,a[l].v);p=getfa(c[j].x); q=getfa(c[j].y);ans[c[j].k]=(p==q && fu[p]==c[j].u && fv[p]==c[j].v);for (; tot; tot--){p=h[tot].x; q=h[tot].y; fa[p]=p;fu[q]=h[tot].u; fv[q]=h[tot].v; sz[q]=h[tot].k;}}}for (i=1; i<=cnt; i++) puts(ans[i]?"Yes":"No");return 0;}

by lych

2016.4.21