分析
可以发现第一列和最后一列永远不会被删除,于是我们可以想到维护前后缀最小生成树,但是直接维护的话显然时间空间两爆炸。(通过上网找题解)可以发现我们关心的只是最左边和最右边两列,而不关心内部的连边情况。所以我们可以仅维护这两列的节点在最小生成树上形成的虚树,边权是对应链上最大的边权,合并时对两棵虚树上的所有边再跑一遍最小生成树就好了。
由于虚树的大小是(O(n))级别的,所以该算法的时间复杂度为(O(n(m+q) log n))。
(这代码写起来有点恶心。)
代码
#include <bits/stdc++.h>
#define rin(i,a,b) for(int i=(a);i<=(b);++i)
#define irin(i,a,b) for(int i=(a);i>=(b);--i)
#define trav(i,a) for(int i=head[a];i;i=e[i].nxt)
#define Size(a) (int)a.size()
#define pb push_back
#define mkpr std::make_pair
#define fi first
#define se second
#define lowbit(a) ((a)&(-(a)))
typedef long long LL;
using std::cerr;
using std::endl;
inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int MAXN=105;
const int MAXM=10005;
int n,m,q;
unsigned int SA, SB, SC;int lim;
int getweight() {
SA ^= SA << 16;
SA ^= SA >> 5;
SA ^= SA << 1;
unsigned int t = SA;
SA = SB;
SB = SC;
SC ^= t ^ SA;
return SC % lim + 1;
}
struct mst_edge{
int u,v,w;
inline friend bool operator < (mst_edge x,mst_edge y){
return x.w<y.w;
}
};
struct mst{
LL sum;
std::vector<int> poi;
std::vector<mst_edge> edg;
inline void init(){sum=0;poi.clear();edg.clear();}
};
mst pre[MAXM],suf[MAXM];
std::vector<mst_edge> hng[MAXM],shu[MAXM],ext;
inline int calc_id(int x,int y){
return (x-1)*m+y;
}
int dsu[MAXN*MAXM];
int ecnt,head[MAXN*MAXM];
bool mark[MAXN*MAXM],havem[MAXN*MAXM];
std::vector<int> po,retp;
std::vector<mst_edge> ee,rete;
struct Edge{
int to,nxt,w;
}e[MAXN*MAXM*4];
int getf(int x){
return dsu[x]==x?x:dsu[x]=getf(dsu[x]);
}
inline bool merge_dsu(int x,int y){
x=getf(x),y=getf(y);
if(x!=y){
dsu[y]=x;
return true;
}
return false;
}
inline void add_edge(int bg,int ed,int val){
++ecnt;
e[ecnt].to=ed;
e[ecnt].nxt=head[bg];
e[ecnt].w=val;
head[bg]=ecnt;
}
void dfs1(int x,int pre){
havem[x]|=mark[x];
trav(i,x){
int ver=e[i].to;
if(ver==pre)continue;
dfs1(ver,x);
havem[x]|=havem[ver];
}
}
void dfs2(int x,int pre,int las,int maxw){
int flag=0;
trav(i,x){
int ver=e[i].to;
if(ver==pre)continue;
if(havem[ver])++flag;
}
if(flag>1)mark[x]=true;
if(mark[x]){
retp.pb(x);
if(las)rete.pb((mst_edge){las,x,maxw});
}
trav(i,x){
int ver=e[i].to;
if(ver==pre)continue;
if(mark[x])dfs2(ver,x,x,e[i].w);
else dfs2(ver,x,las,std::max(maxw,e[i].w));
}
}
mst merge(mst L,mst R,int bg){
mst ret;ret.init();
ret.sum=L.sum+R.sum;
po.clear();ee.clear();
retp.clear();rete.clear();ecnt=0;
rin(i,0,Size(L.poi)-1)po.pb(L.poi[i]);
rin(i,0,Size(R.poi)-1)po.pb(R.poi[i]);
rin(i,0,Size(L.edg)-1){
ret.sum-=L.edg[i].w;
ee.pb(L.edg[i]);
}
rin(i,0,Size(R.edg)-1){
ret.sum-=R.edg[i].w;
ee.pb(R.edg[i]);
}
rin(i,0,Size(ext)-1)ee.pb(ext[i]);
rin(i,0,Size(po)-1){
dsu[po[i]]=po[i];
head[po[i]]=0;
mark[po[i]]=havem[po[i]]=false;
}
std::sort(ee.begin(),ee.end());
int temp=0;
rin(i,0,Size(ee)-1){
if(merge_dsu(ee[i].u,ee[i].v)){
++temp;
ret.sum+=ee[i].w;
add_edge(ee[i].u,ee[i].v,ee[i].w);
add_edge(ee[i].v,ee[i].u,ee[i].w);
if(temp==Size(po)-1)break;
}
}
rin(i,1,n)mark[calc_id(i,bg)]=true;
if(bg==1)rin(i,0,Size(R.poi)-1)mark[R.poi[i]]=true;
else rin(i,0,Size(L.poi)-1)mark[L.poi[i]]=true;
dfs1(po[0],0);dfs2(po[0],0,0,0);
ret.poi=retp,ret.edg=rete;
return ret;
}
int main(){
n=read(),m=read(),SA=read(),SB=read(),SC=read(),lim=read();
rin(i,1,n)rin(j,1,m){
int w=getweight();
if(j<m)hng[j].pb((mst_edge){calc_id(i,j),calc_id(i,j+1),w});
else hng[j].pb((mst_edge){calc_id(i,j),calc_id(i,1),w});
}
rin(i,1,n-1)rin(j,1,m){
int w=getweight();
shu[j].pb((mst_edge){calc_id(i,j),calc_id(i+1,j),w});
}
rin(i,1,n)pre[1].poi.pb(calc_id(i,1));
rin(i,0,Size(shu[1])-1){
pre[1].sum+=shu[1][i].w;
pre[1].edg.pb(shu[1][i]);
}
rin(i,2,m){
ext.clear();
rin(j,1,n)pre[i].poi.pb(calc_id(j,i));
rin(j,0,Size(shu[i])-1){
pre[i].sum+=shu[i][j].w;
pre[i].edg.pb(shu[i][j]);
}
rin(j,0,Size(hng[i-1])-1)ext.pb(hng[i-1][j]);
pre[i]=merge(pre[i-1],pre[i],1);
}
rin(i,1,n)suf[m].poi.pb(calc_id(i,m));
rin(i,0,Size(shu[m])-1){
suf[m].sum+=shu[m][i].w;
suf[m].edg.pb(shu[m][i]);
}
irin(i,m-1,1){
ext.clear();
rin(j,1,n)suf[i].poi.pb(calc_id(j,i));
rin(j,0,Size(shu[i])-1){
suf[i].sum+=shu[i][j].w;
suf[i].edg.pb(shu[i][j]);
}
rin(j,0,Size(hng[i])-1)ext.pb(hng[i][j]);
suf[i]=merge(suf[i],suf[i+1],m);
}
ext.clear();
rin(i,0,Size(hng[m])-1)ext.pb(hng[m][i]);
q=read();
while(q--){
int l=read(),r=read();
printf("%lld
",merge(pre[l-1],suf[r+1],1).sum);
}
return 0;
}