【BZOJ 3545】【ONTAK 2010】Peaks & 【BZOJ 3551】【ONTAK 2010】Peaks加强版 Kruskal重构树

时间:2024-01-19 08:44:38

sunshine的A题我竟然调了一周!!!

把循环dfs改成一个dfs就可以,,,我也不知道为什么这样就不会RE,但它却是A了,,,

这周我一直在调这个题,总结一下智障错误:

1.倍增的范围设成了n而不是n*2-1,,,

2.重构树的顶点是n*2-1,而我一开始设成了n,,,

3.define里的for3和for4的i--打成i++,,,,,,,,,,,,

4.dfs爆栈了,找CA爷问的编译命令里手动扩栈,真是愚蠢的问题,,,,

比赛时绝不会有太多时间,在这么犯逗就得滚粗了QAQ

3545:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define for1(i,a,n) for(int i=(a);i<=(n);i++)
#define for2(i,a,n) for(int i=(a);i<(n);i++)
#define for3(i,a,n) for(int i=(a);i>=(n);i--)
#define for4(i,a,n) for(int i=(a);i>(n);i--)
#define CC(i,a) memset(i,a,sizeof(i));
#define read(x) x=getint()
using namespace std;
inline const int getint(){char c=getchar();int k=1,r=0;for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
inline const int max(const int &a,const int &b){return a>b?a:b;}
inline const int min(const int &a,const int &b){return a<b?a:b;}
const int N=1E5+10;
const int M=5*1E5+10;
struct node{int x,y,z;}E[M];
struct snnn{int l,r,s;}T[N*30];
int n,m,fa[N<<1],num[N<<1],HH[N],id[N],H[N],cnt,f[N<<1][20];
int root[N],L[N<<1],R[N<<1],lch[N<<1],rch[N<<1],ST[N];
inline bool cmp(node X,node Y){return X.z<Y.z;}
inline bool cmp2(int X,int Y){return HH[X]<HH[Y];}
inline void init(){CC(fa,0);CC(num,0);CC(HH,0);CC(id,0);CC(H,0);CC(f,0);CC(root,0);CC(L,0);CC(R,0);CC(lch,0);CC(rch,0);CC(ST,0);}
inline int find(int x){
if (x==fa[x]) return x;
else{fa[x]=find(fa[x]);return fa[x];}
}
inline void LCA(int dd){
for1(j,1,19)
for1(i,1,dd)
f[i][j]=f[f[i][j-1]][j-1];
}
inline void dfs(int x){
if (lch[x]&&rch[x]){
L[x]=cnt+1;
dfs(lch[x]);
dfs(rch[x]);
R[x]=cnt;
}else{
cnt++; ST[cnt]=x;
L[x]=cnt; R[x]=cnt;
}
}
inline void update(int l,int r,int &pos,int key){
T[++cnt]=T[pos]; pos=cnt; T[pos].s++;
if (l==r) return;
int mid=(l+r)>>1;
if (key<=mid) update(l,mid,T[pos].l,key); else update(mid+1,r,T[pos].r,key);
}
inline int LCA_find(int x,int y){
for3(i,19,0)
if ((f[x][i]!=0)&&(num[f[x][i]]<=y))
x=f[x][i];
return x;
}
inline int query(int LL,int RR,int key){
int mid,s,x=root[LL],y=root[RR],l=1,r=n;
while (l<r){
mid=(l+r)>>1; s=T[T[y].l].s-T[T[x].l].s;
if (key<=s) r=mid,x=T[x].l,y=T[y].l;
else l=mid+1,key-=s,x=T[x].r,y=T[y].r;
}return l;
}
int main(){
init();
read(n); read(m); int Q; read(Q);
for1(i,1,n) read(HH[i]),id[i]=i;
sort(id+1,id+n+1,cmp2);
for1(i,1,n) H[id[i]]=i;
for1(i,1,m){read(E[i].x);read(E[i].y);read(E[i].z);}
sort(E+1,E+m+1,cmp);
cnt=n+1;
for2(i,1,n<<1) fa[i]=i;
for1(i,1,m){
int fx=find(E[i].x),fy=find(E[i].y);
if (fx!=fy){
fa[fx]=cnt; fa[fy]=cnt;
f[fx][0]=cnt; f[fy][0]=cnt;
lch[cnt]=fx; rch[cnt]=fy;
num[cnt]=E[i].z;
cnt++;
if (cnt>((n<<1)-1)) break;
}
}
int dd=cnt-1;
LCA(dd);
cnt=0;
//for2(i,1,n<<1) if (L[i]==0) dfs(find(i));
dfs(dd);
cnt=0;
for1(i,1,n){
root[i]=root[i-1];
update(1,n,root[i],H[ST[i]]);
}
int a,b,c,la=0;
for1(i,1,Q){
read(a); read(b); read(c);
//a^=la; b^=la; c^=la;
int rt=LCA_find(a,b),nn;
if (R[rt]-L[rt]+1<c) nn=-1;
else nn=HH[id[query(L[rt]-1,R[rt],R[rt]-L[rt]+2-c)]],la=nn;
printf("%d\n",nn);
}
return 0;
}

3551:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define for1(i,a,n) for(int i=(a);i<=(n);i++)
#define for2(i,a,n) for(int i=(a);i<(n);i++)
#define for3(i,a,n) for(int i=(a);i>=(n);i--)
#define for4(i,a,n) for(int i=(a);i>(n);i--)
#define CC(i,a) memset(i,a,sizeof(i));
#define read(x) x=getint()
using namespace std;
inline const int getint(){char c=getchar();int k=1,r=0;for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
inline const int max(const int &a,const int &b){return a>b?a:b;}
inline const int min(const int &a,const int &b){return a<b?a:b;}
const int N=1E5+10;
const int M=5*1E5+10;
struct node{int x,y,z;}E[M];
struct snnn{int l,r,s;}T[N*30];
int n,m,fa[N<<1],num[N<<1],HH[N],id[N],H[N],cnt,f[N<<1][20];
int root[N],L[N<<1],R[N<<1],lch[N<<1],rch[N<<1],ST[N];
inline bool cmp(node X,node Y){return X.z<Y.z;}
inline bool cmp2(int X,int Y){return HH[X]<HH[Y];}
inline void init(){CC(fa,0);CC(num,0);CC(HH,0);CC(id,0);CC(H,0);CC(f,0);CC(root,0);CC(L,0);CC(R,0);CC(lch,0);CC(rch,0);CC(ST,0);}
inline int find(int x){
if (x==fa[x]) return x;
else{fa[x]=find(fa[x]);return fa[x];}
}
inline void LCA(int dd){
for1(j,1,19)
for1(i,1,dd)
f[i][j]=f[f[i][j-1]][j-1];
}
inline void dfs(int x){
if (lch[x]&&rch[x]){
L[x]=cnt+1;
dfs(lch[x]);
dfs(rch[x]);
R[x]=cnt;
}else{
cnt++; ST[cnt]=x;
L[x]=cnt; R[x]=cnt;
}
}
inline void update(int l,int r,int &pos,int key){
T[++cnt]=T[pos]; pos=cnt; T[pos].s++;
if (l==r) return;
int mid=(l+r)>>1;
if (key<=mid) update(l,mid,T[pos].l,key); else update(mid+1,r,T[pos].r,key);
}
inline int LCA_find(int x,int y){
for3(i,19,0)
if ((f[x][i]!=0)&&(num[f[x][i]]<=y))
x=f[x][i];
return x;
}
inline int query(int LL,int RR,int key){
int mid,s,x=root[LL],y=root[RR],l=1,r=n;
while (l<r){
mid=(l+r)>>1; s=T[T[y].l].s-T[T[x].l].s;
if (key<=s) r=mid,x=T[x].l,y=T[y].l;
else l=mid+1,key-=s,x=T[x].r,y=T[y].r;
}return l;
}
int main(){
init();
read(n); read(m); int Q; read(Q);
for1(i,1,n) read(HH[i]),id[i]=i;
sort(id+1,id+n+1,cmp2);
for1(i,1,n) H[id[i]]=i;
for1(i,1,m){read(E[i].x);read(E[i].y);read(E[i].z);}
sort(E+1,E+m+1,cmp);
cnt=n+1;
for2(i,1,n<<1) fa[i]=i;
for1(i,1,m){
int fx=find(E[i].x),fy=find(E[i].y);
if (fx!=fy){
fa[fx]=cnt; fa[fy]=cnt;
f[fx][0]=cnt; f[fy][0]=cnt;
lch[cnt]=fx; rch[cnt]=fy;
num[cnt]=E[i].z;
cnt++;
if (cnt>((n<<1)-1)) break;
}
}
int dd=cnt-1;
LCA(dd);
cnt=0;
//for2(i,1,n<<1) if (L[i]==0) dfs(find(i));
dfs(dd);
cnt=0;
for1(i,1,n){
root[i]=root[i-1];
update(1,n,root[i],H[ST[i]]);
}
int a,b,c,la=0;
for1(i,1,Q){
read(a); read(b); read(c);
a^=la; b^=la; c^=la;
int rt=LCA_find(a,b),nn;
if (R[rt]-L[rt]+1<c) nn=-1,la=0;
else nn=HH[id[query(L[rt]-1,R[rt],R[rt]-L[rt]+2-c)]],la=nn;
printf("%d\n",nn);
}
return 0;
}

然后就可以了O(∩_∩)O~~