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[SDOI2011]染色 题解

题目大意:

  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)

思路:

  树剖之后,维护其两端的颜色、答案和标记即可。

代码:

#include<cstdio>#include<iostream>#define N 100001using namespace std;int n,m,cnt,dfn,vis[N],head[N],to[N<<1],next[N<<1],deep[N],size[N],top[N],id[N],v[N],pa[N],lazy[N<<2],lc[N<<2],rc[N<<2],sum[N<<2];void ins(int x,int y){     to[++cnt]=y,next[cnt]=head[x],head[x]=cnt;}void dfs1(int x){    size[x]=vis[x]=1;    for (int i=head[x];i;i=next[i])        if (!vis[to[i]])        {            deep[to[i]]=deep[x]+1;            pa[to[i]]=x;            dfs1(to[i]);            size[x]+=size[to[i]];        }}void dfs2(int x,int chain){    int k=0,i;    id[x]=++dfn;    top[x]=chain;    for (i=head[x];i;i=next[i])        if (deep[to[i]]>deep[x] && size[to[i]]>size[k]) k=to[i];    if (!k) return;    dfs2(k,chain);    for (i=head[x];i;i=next[i])        if (deep[to[i]]>deep[x] && to[i]!=k) dfs2(to[i],to[i]);}void build(int l,int r,int cur){    sum[cur]=1,lazy[cur]=-1;    if(l==r)return;    int mid=l+r>>1;    build(l,mid,cur<<1),build(mid+1,r,cur<<1|1);}void update(int k){    lc[k]=lc[k<<1],rc[k]=rc[k<<1|1];    if (rc[k<<1]^lc[k<<1|1]) sum[k]=sum[k<<1]+sum[k<<1|1];    else sum[k]=sum[k<<1]+sum[k<<1|1]-1;}void pushdown(int l,int r,int k){    int tmp=lazy[k]; lazy[k]=-1;    if (tmp==-1 || l==r) return;    sum[k<<1]=sum[k<<1|1]=1;    lazy[k<<1]=lazy[k<<1|1]=tmp;    lc[k<<1]=rc[k<<1]=tmp;    lc[k<<1|1]=rc[k<<1|1]=tmp;}void change(int L,int R,int l,int r,int cur,int val){    if (l==L && r==R) { lc[cur]=rc[cur]=val; sum[cur]=1; lazy[cur]=val; return; }    int mid=L+R>>1; pushdown(L,R,cur);    if (r<=mid) change(L,mid,l,r,cur<<1,val);    else if (l>mid) change(mid+1,R,l,r,cur<<1|1,val);         else change(L,mid,l,mid,cur<<1,val),change(mid+1,R,mid+1,r,cur<<1|1,val);    update(cur);}int ask(int L,int R,int l,int r,int cur){    if (l==L && r==R) return sum[cur];    int mid=L+R>>1; pushdown(L,R,cur);    if (r<=mid) return ask(L,mid,l,r,cur<<1);    else if (l>mid) return ask(mid+1,R,l,r,cur<<1|1);         else         {              int tmp=1;              if (rc[cur<<1]^lc[cur<<1|1]) tmp=0;              return ask(L,mid,l,mid,cur<<1)+ask(mid+1,R,mid+1,r,cur<<1|1)-tmp;         }}int getc(int l,int r,int cur,int x){    if (l==r) return lc[cur];    int mid=l+r>>1; pushdown(l,r,cur);    if (x<=mid) return getc(l,mid,cur<<1,x);    else return getc(mid+1,r,cur<<1|1,x);}int solvesum(int x,int y){    int sum=0;    for (;top[x]!=top[y];x=pa[top[x]])    {        if (deep[top[x]]<deep[top[y]]) swap(x,y);        sum+=ask(1,n,id[top[x]],id[x],1);        if (getc(1,n,1,id[top[x]])==getc(1,n,1,id[pa[top[x]]])) sum--;    }    if (deep[x]>deep[y]) swap(x,y);    return sum+=ask(1,n,id[x],id[y],1);}void solvechange(int x,int y,int val){    for (;top[x]!=top[y];x=pa[top[x]])    {        if (deep[top[x]]<deep[top[y]]) swap(x,y);        change(1,n,id[top[x]],id[x],1,val);    }    if (deep[x]>deep[y]) swap(x,y);    change(1,n,id[x],id[y],1,val);}int main(){    int i,a,b,c;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)scanf("%d",&v[i]);    for(i=1;i<n;i++) scanf("%d%d",&a,&b),ins(a,b),ins(b,a);    dfs1(1),dfs2(1,1),build(1,n,1);    for(i=1;i<=n;i++) change(1,n,id[i],id[i],1,v[i]);    for(i=1;i<=m;i++)    {        char ch[10];        scanf("%s",ch);        if(ch[0]==Q)        {            scanf("%d%d",&a,&b);            printf("%d\n",solvesum(a,b));        }        else        {            scanf("%d%d%d",&a,&b,&c);            solvechange(a,b,c);        }    }    return 0;}

 

[SDOI2011]染色 题解