Bzoj 2243: [SDOI2011]染色 树链剖分,LCT,动态树
2016-03-23 15:59
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2243: [SDOI2011]染色
Time Limit: 20 Sec Memory Limit: 512 MBSubmit: 5020 Solved: 1872
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Description
给定一棵有n个节点的无根树和m个操作,操作有2类:1、将节点a到节点b路径上所有点都染成颜色c;
2、询问节点a到节点b路径上的颜色段数量(连续相同颜色被认为是同一段),如“112221”由3段组成:“11”、“222”和“1”。
请你写一个程序依次完成这m个操作。
Input
第一行包含2个整数n和m,分别表示节点数和操作数;第二行包含n个正整数表示n个节点的初始颜色
下面 行每行包含两个整数x和y,表示x和y之间有一条无向边。
下面 行每行描述一个操作:
“C a b c”表示这是一个染色操作,把节点a到节点b路径上所有点(包括a和b)都染成颜色c;
“Q a b”表示这是一个询问操作,询问节点a到节点b(包括a和b)路径上的颜色段数量。
Output
对于每个询问操作,输出一行答案。Sample Input
6 52 2 1 2 1 1
1 2
1 3
2 4
2 5
2 6
Q 3 5
C 2 1 1
Q 3 5
C 5 1 2
Q 3 5
Sample Output
31
2
HINT
数N<=10^5,操作数M<=10^5,所有的颜色C为整数且在[0, 10^9]之间。Source
第一轮day1题解:
树链剖分处理一下连接点之间的颜色是否相同即可。。。
线段树中记录 左右颜色,分成段数即可。。。
其实LCT做这道题也是不错的呦!!!(两个程序都挂上吧。。。)
对比:
LCT:
9088 kb | 17756 ms |
29944 kb | 8652 ms |
#include<bits/stdc++.h> using namespace std; #define MAXN 100010 struct node { int begin,end,next; }edge[MAXN*2]; struct NODE { int left,right,lc,rc,color,tag,sum; }tree[MAXN*5]; int cnt,Head[MAXN],size[MAXN],deep[MAXN],P[MAXN][17],pos[MAXN],belong[MAXN],c[MAXN],SIZE,n; bool vis[MAXN]; void addedge(int bb,int ee) { edge[++cnt].begin=bb;edge[cnt].end=ee;edge[cnt].next=Head[bb];Head[bb]=cnt; } void addedge1(int bb,int ee) { addedge(bb,ee);addedge(ee,bb); } int read() { int s=0,fh=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} return s*fh; } void dfs1(int u) { int i,v; size[u]=1;vis[u]=true; for(i=Head[u];i!=-1;i=edge[i].next) { v=edge[i].end; if(vis[v]==false) { deep[v]=deep[u]+1; P[v][0]=u; dfs1(v); size[u]+=size[v]; } } } void Ycl() { int i,j; for(j=1;(1<<j)<=n;j++) { for(i=1;i<=n;i++) { if(P[i][j-1]!=-1)P[i][j]=P[P[i][j-1]][j-1]; } } } void dfs2(int u,int chain) { int k=0,i,v; pos[u]=++SIZE;belong[u]=chain; for(i=Head[u];i!=-1;i=edge[i].next) { v=edge[i].end; if(deep[v]>deep[u]&&size[v]>size[k])k=v; } if(k==0)return; dfs2(k,chain); for(i=Head[u];i!=-1;i=edge[i].next) { v=edge[i].end; if(deep[v]>deep[u]&&v!=k)dfs2(v,v); } } int LCA(int x,int y) { int i,j; if(deep[x]<deep[y])swap(x,y); for(i=0;(1<<i)<=deep[x];i++);i--; for(j=i;j>=0;j--)if(deep[x]-(1<<j)>=deep[y])x=P[x][j]; if(x==y)return x; for(j=i;j>=0;j--) { if(P[x][j]!=-1&&P[x][j]!=P[y][j]) { x=P[x][j]; y=P[y][j]; } } return P[x][0]; } void Pushup(int k) { int l=k*2,r=k*2+1; tree[k].lc=tree[l].lc; tree[k].rc=tree[r].rc; if(tree[l].rc==tree[r].lc)tree[k].sum=tree[l].sum+tree[r].sum-1; else tree[k].sum=tree[l].sum+tree[r].sum; } void Pushdown(int k) { int l=k*2,r=k*2+1; if(tree[k].tag!=-1) { tree[l].tag=tree[k].tag; tree[r].tag=tree[k].tag; tree[l].lc=tree[l].rc=tree[l].color=tree[k].tag; tree[r].lc=tree[r].rc=tree[r].color=tree[k].tag; tree[l].sum=tree[r].sum=1; tree[k].tag=-1; } } void Build(int k,int l,int r) { tree[k].left=l;tree[k].right=r;tree[k].tag=-1; if(l==r)return; int mid=(l+r)/2; Build(k*2,l,mid);Build(k*2+1,mid+1,r); } void Change(int k,int l,int r,int C) { if(l<=tree[k].left&&tree[k].right<=r){tree[k].lc=tree[k].rc=tree[k].color=tree[k].tag=C;tree[k].sum=1;return;} Pushdown(k); int mid=(tree[k].left+tree[k].right)/2; if(r<=mid)Change(k*2,l,r,C); else if(l>mid)Change(k*2+1,l,r,C); else {Change(k*2,l,mid,C);Change(k*2+1,mid+1,r,C);} Pushup(k); } void Solve_change(int x,int f,int C) { while(belong[x]!=belong[f]) { Change(1,pos[belong[x]],pos[x],C); x=P[belong[x]][0]; } Change(1,pos[f],pos[x],C); } int Query_sum(int k,int l,int r) { if(l<=tree[k].left&&tree[k].right<=r)return tree[k].sum; Pushdown(k); int mid=(tree[k].left+tree[k].right)/2; if(r<=mid)return Query_sum(k*2,l,r); else if(l>mid)return Query_sum(k*2+1,l,r); //else return Query_sum(k*2,l,mid)+Query_sum(k*2+1,mid+1,r); else { if(tree[k*2].rc==tree[k*2+1].lc)return Query_sum(k*2,l,mid)+Query_sum(k*2+1,mid+1,r)-1; else return Query_sum(k*2,l,mid)+Query_sum(k*2+1,mid+1,r); } //Pushup(k); } int get_color(int k,int lr) { if(tree[k].left==tree[k].right)return tree[k].color; Pushdown(k); int mid=(tree[k].left+tree[k].right)/2; if(lr<=mid)return get_color(k*2,lr); else if(lr>mid)return get_color(k*2+1,lr); } int Solve_sum(int x,int f) { int sum=0; while(belong[x]!=belong[f]) { sum+=Query_sum(1,pos[belong[x]],pos[x]); if(get_color(1,pos[belong[x]])==get_color(1,pos[P[belong[x]][0]]))sum--; x=P[belong[x]][0]; } sum+=Query_sum(1,pos[f],pos[x]); return sum; } int main() { int m,i,bb,ee,A,B,C,lca; char zs[2]; n=read();m=read(); for(i=1;i<=n;i++)c[i]=read(); memset(Head,-1,sizeof(Head));cnt=1; for(i=1;i<n;i++){bb=read();ee=read();addedge1(bb,ee);} memset(P,-1,sizeof(P));SIZE=0; dfs1(1);Ycl(); dfs2(1,1); Build(1,1,n); for(i=1;i<=n;i++)Change(1,pos[i],pos[i],c[i]); for(i=1;i<=m;i++) { scanf("%s",zs); if(zs[0]=='C') { A=read();B=read();C=read(); lca=LCA(A,B); Solve_change(A,lca,C);Solve_change(B,lca,C); } else { A=read();B=read(); lca=LCA(A,B); printf("%d\n",Solve_sum(A,lca)+Solve_sum(B,lca)-1); } } fclose(stdin); fclose(stdout); return 0; }
LCT(代码略丑):
#include<bits/stdc++.h> using namespace std; #define LL long long #define MAXN 100010 #define INF 1e9 struct node { LL left,right,color,tag1,lc,rc,sum; }tree[MAXN]; LL rev[MAXN],father[MAXN],Stack[MAXN]; LL read() { LL s=0,fh=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} return s*fh; } LL isroot(LL x) { return tree[father[x]].left!=x&&tree[father[x]].right!=x; } void pushdown(LL x) { LL l=tree[x].left,r=tree[x].right; if(rev[x]!=0) { rev[x]^=1;rev[l]^=1;rev[r]^=1; swap(tree[x].left,tree[x].right); swap(tree[x].lc,tree[x].rc); //tree[x].lc=tree[r].lc;tree[x].rc=tree[l].rc; //swap(tree[l].color,tree[r].color); //swap(tree[l].sum,tree[r].sum); } if(tree[x].tag1!=-1) { //tree[l].lc=tree[l].rc=tree[x].tag1; //tree[r].lc=tree[r].rc=tree[x].tag1; /*tree[l].color=*/tree[x].color=tree[x].lc=tree[x].rc=tree[x].tag1; tree[l].tag1=tree[r].tag1=tree[x].tag1; tree[x].sum=1; tree[x].tag1=-1; } } void pushup(LL x) { LL l=tree[x].left,r=tree[x].right; //if(tree[l].rc==tree[r].lc) //{ if(l!=0)pushdown(l); if(r!=0)pushdown(r); //if(l!=0)lr+=tree[l].sum; //if(r!=0)lr+=tree[r].sum; tree[x].sum=tree[l].sum+tree[r].sum+1; if(l!=0&&tree[l].rc==tree[x].color)tree[x].sum--; if(r!=0&&tree[r].lc==tree[x].color)tree[x].sum--; //tree[x].sum=lr; tree[x].lc=tree[x].rc=tree[x].color; if(l!=0)tree[x].lc=tree[l].lc; if(r!=0)tree[x].rc=tree[r].rc; //tree[x].sum=tree[l].sum+tree[r].sum-1; //tree[x].lc=tree[l].lc; //tree[x].rc=tree[r].rc; //} //else //{ //tree[x].sum=tree[l].sum+tree[r].sum; //tree[x].lc=tree[l].lc; //tree[x].rc=tree[r].rc; //if(l!=0)lr+=tree[l].sum,tree[x].lc=tree[l].lc; //if(r!=0)lr+=tree[r].sum,tree[x].rc=tree[r].rc; //if(lr!=0)tree[x].sum=lr+1; //} } void rotate(LL x) { LL y=father[x],z=father[y]; if(!isroot(y)) { if(tree[z].left==y)tree[z].left=x; else tree[z].right=x; } if(tree[y].left==x) { father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y; } else { father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y; } pushup(y);pushup(x); } void splay(LL x) { LL top=0,i,y,z;Stack[++top]=x; for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i]; for(i=top;i>=1;i--)pushdown(Stack[i]); while(!isroot(x)) { y=father[x];z=father[y]; if(!isroot(y)) { if((tree[y].left==x)^(tree[z].left==y))rotate(x); else rotate(y); } rotate(x); } } void access(LL x) { LL last=0; while(x!=0) { splay(x); tree[x].right=last;pushup(x); last=x;x=father[x]; } } void makeroot(LL x) { access(x);splay(x);rev[x]^=1; } void link(LL u,LL v) { makeroot(u);father[u]=v;splay(u); } void cut(LL u,LL v) { makeroot(u);access(v);splay(v);father[u]=tree[v].left=0; } LL findroot(LL x) { access(x);splay(x); while(tree[x].left!=0)x=tree[x].left; return x; } int main() { LL i,x,y,n,m,a,b,c; char fh[2]; n=read();m=read(); tree[0].lc=tree[0].rc=-INF; for(i=1;i<=n;i++)tree[i].color=tree[i].lc=tree[i].rc=read(),tree[i].sum=1,tree[i].tag1=-1; for(i=1;i<n;i++) { x=read();y=read(); link(x,y); } for(i=1;i<=m;i++) { scanf("\n%s",fh); if(fh[0]=='C') { a=read();b=read();c=read(); makeroot(a);access(b);splay(b); tree[b].color=tree[b].lc=tree[b].rc=c;tree[b].tag1=c; tree[b].sum=1; } else { a=read();b=read(); makeroot(a);access(b);splay(b); printf("%d\n",tree[b].sum); } } return 0; }
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