bzoj1759: Let there be rainbows!

传送门
显然这道题是可以用树剖水过去的。

#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#define N 200005using namespace std;struct edge{int to,next;}e[N*2];int head[N],sz[N],fa[N][20],dep[N],pos[N],top[N];int tot,cnt,n,m,x,y,z,bin[25],ans[8];struct node{int tag,s[8];}t[N*4];void add(int x,int y){    e[++tot]=(edge){y,head[x]};    head[x]=tot;}void dfs1(int x){    sz[x]=1;    for (int i=1;i<=19;i++)        fa[x][i]=fa[fa[x][i-1]][i-1];    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]){            dep[e[i].to]=dep[x]+1;            fa[e[i].to][0]=x;            dfs1(e[i].to);            sz[x]+=sz[e[i].to];        }}void dfs2(int x,int tp){    pos[x]=++cnt; top[x]=tp;    int k=0;    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]&&sz[e[i].to]>sz[k])            k=e[i].to;    if (k) dfs2(k,tp);    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]&&e[i].to!=k)            dfs2(e[i].to,e[i].to);}int lca(int x,int y){    if (dep[x]<dep[y]) swap(x,y);    int tmp=dep[x]-dep[y];    for (int i=19;i>=0;i--)        if (tmp&bin[i]) x=fa[x][i];    for (int i=19;i>=0;i--)        if (fa[x][i]!=fa[y][i])            x=fa[x][i],y=fa[y][i];    return (x==y)?x:fa[x][0];}void setcolor(int k,int c,int sz){    memset(t[k].s,0,sizeof(t[k].s));    t[k].s[c]=sz;    t[k].tag=c;}void pushdown(int k,int l,int r){    if (!t[k].tag||l==r) return;    int mid=(l+r)/2;    setcolor(k*2,t[k].tag,mid-l+1);    setcolor(k*2+1,t[k].tag,r-mid);    t[k].tag=0;}void build(int k,int l,int r){    t[k].s[0]=r-l+1;    if (l==r) return;    int mid=(l+r)/2;    build(k*2,l,mid);    build(k*2+1,mid+1,r);}int ask(int k,int l,int r,int x,int y,int c){    int mid=(l+r)/2,ans;    pushdown(k,l,r);    if (l==x&&r==y){        ans=r-l+1-t[k].s[c];        setcolor(k,c,r-l+1);        return ans;    }    if (y<=mid) ans=ask(k*2,l,mid,x,y,c);    else if (x>mid) ans=ask(k*2+1,mid+1,r,x,y,c);    else ans=ask(k*2,l,mid,x,mid,c)+ask(k*2+1,mid+1,r,mid+1,y,c);    for (int i=0;i<8;i++) t[k].s[i]=t[k*2].s[i]+t[k*2+1].s[i];    return ans;}int ask(int x,int f,int z){    int ans=0;    for (;top[x]!=top[f];x=fa[top[x]][0])        ans+=ask(1,1,n,pos[top[x]],pos[x],z);    if (pos[f]+1<=pos[x])        ans+=ask(1,1,n,pos[f]+1,pos[x],z);    return ans;}int main(){    bin[0]=1;    for (int i=1;i<=20;i++)        bin[i]=bin[i-1]*2;    scanf("%d",&n);    for (int i=1;i<n;i++){        scanf("%d%d",&x,&y);        add(x,y); add(y,x);    }    dfs1(1);    dfs2(1,1);    build(1,1,n);    scanf("%d",&m);    while (m--){        scanf("%d%d%d",&x,&y,&z);        int f=lca(x,y);        ans[z]+=ask(x,f,z);        ans[z]+=ask(y,f,z);    }    for (int i=1;i<=7;i++)        printf("%d\n",ans[i]);}//TLE代码

然后，这道题卡内存+卡时限。
首先加了一下读入优化。
然后发现这道题我们只要维护7种颜色而非八种。
然后就驶过去辣。

#include<cmath>#include<ctime>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#define N 200005using namespace std;struct edge{int to,next;}e[N*2];int head[N],sz[N],fa[N][19],dep[N],pos[N],top[N];int tot,cnt,n,m,x,y,z,bin[25];struct node{int tag,s[7];}t[N*4];long long ans[8];inline int read(){    int x=0;    char ch=getchar();    for (;ch<'0'||ch>'9';ch=getchar());    for (;ch>='0'&&ch<='9';ch=getchar())        x=x*10-48+ch;    return x;}void add(int x,int y){    e[++tot]=(edge){y,head[x]};    head[x]=tot;}void dfs1(int x){    sz[x]=1;    for (int i=1;i<=18;++i)        fa[x][i]=fa[fa[x][i-1]][i-1];    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]){            dep[e[i].to]=dep[x]+1;            fa[e[i].to][0]=x;            dfs1(e[i].to);            sz[x]+=sz[e[i].to];        }}void dfs2(int x,int tp){    pos[x]=++cnt; top[x]=tp;    int k=0;    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]&&sz[e[i].to]>sz[k])            k=e[i].to;    if (k) dfs2(k,tp);    for (int i=head[x];i;i=e[i].next)        if (e[i].to!=fa[x][0]&&e[i].to!=k)            dfs2(e[i].to,e[i].to);}int lca(int x,int y){    if (dep[x]<dep[y]) swap(x,y);    int tmp=dep[x]-dep[y];    for (int i=18;i>=0;--i)        if (tmp&bin[i]) x=fa[x][i];    for (int i=18;i>=0;--i)        if (fa[x][i]!=fa[y][i])            x=fa[x][i],y=fa[y][i];    return (x==y)?x:fa[x][0];}void setcolor(int k,int c,int sz){    memset(t[k].s,0,sizeof(t[k].s));    t[k].s[c]=sz;    t[k].tag=c;}void pushdown(int k,int l,int r){    if (t[k].tag==-1) return;    int mid=(l+r)/2;    setcolor(k*2,t[k].tag,mid-l+1);    setcolor(k*2+1,t[k].tag,r-mid);    t[k].tag=-1;}void build(int k,int l,int r){    t[k].tag=-1;    if (l==r) return;    int mid=(l+r)/2;    build(k*2,l,mid);    build(k*2+1,mid+1,r);}int ask(int k,int l,int r,int x,int y,int c){    int mid=(l+r)/2,ans;    if (l==x&&r==y){        ans=r-l+1-t[k].s[c];        setcolor(k,c,r-l+1);        return ans;    }    pushdown(k,l,r);    if (y<=mid) ans=ask(k*2,l,mid,x,y,c);    else if (x>mid) ans=ask(k*2+1,mid+1,r,x,y,c);    else ans=ask(k*2,l,mid,x,mid,c)+ask(k*2+1,mid+1,r,mid+1,y,c);    for (int i=0;i<7;++i) t[k].s[i]=t[k*2].s[i]+t[k*2+1].s[i];    return ans;}int ask(int x,int f,int z){    int ans=0;    for (;top[x]!=top[f];x=fa[top[x]][0])        ans+=ask(1,1,n,pos[top[x]],pos[x],z);    if (pos[f]+1<=pos[x])        ans+=ask(1,1,n,pos[f]+1,pos[x],z);    return ans;}int main(){    bin[0]=1;    for (int i=1;i<=20;++i)        bin[i]=bin[i-1]*2;    n=read();    for (int i=1;i<n;++i){        x=read(); y=read();        add(x,y); add(y,x);    }    dfs1(1);    dfs2(1,1);    build(1,1,n);    m=read();    while (m--){        x=read(); y=read(); z=read()-1;        int f=lca(x,y);        ans[z]+=ask(x,f,z);        ans[z]+=ask(y,f,z);    }    for (int i=0;i<7;i++)        printf("%d\n",ans[i]);}