UVA 11994 Happy Painting!

最新推荐文章于 2022-08-24 09:05:26 发布

转载最新推荐文章于 2022-08-24 09:05:26 发布 · 83 阅读

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原文链接：http://www.cnblogs.com/staginner/archive/2012/08/23/2652982.html

这篇博客详细介绍了UVA_11994题目的解决方法，主要涉及链接切割树的基本操作以及如何在路径上计数不同颜色的数量。通过使用整数的二进制位来存储颜色种类，实现高效的颜色计数。文章提供了具体的代码实现和算法解释。

UVA_11994

这个题目思维上的障碍比较少，因为实际上都是link-cut-tree的基本的操作，还有一个更为复杂的link-cut-tree的题目——HDU_4010。

在维护路径上边的颜色的数量时，由于颜色的种类只有30个，因此可以用一个整数的30个二进制位来存储颜色的种类。

#include<stdio.h>
#include<string.h>
#define MAXD 50010
int N, M;
struct Splay
{
    int pre, ls, rs, col, set, to, size;
    bool root;
    void pushdown(); void update(); void zig(int x); void zag(int x); void splay(int x);
    void init()
    {
        root = true;
        size = to = pre = ls = rs = col = set = 0;
    }
}sp[MAXD];
inline void same(int x, int c)
{
    if(x) sp[x].to = sp[x].col = c, sp[x].set = 1 << c;
}
inline void Splay::pushdown()
{
    if(to)
    {
        same(ls, to), same(rs, to);
        to = 0;
    }
}
inline void Splay::update()
{
    set = sp[ls].set | sp[rs].set | 1 << col;
    size = sp[ls].size + sp[rs].size + 1;
}
void Splay::zig(int x)
{
    int y = rs, fa = pre;
    rs = sp[y].ls, sp[rs].pre = x;
    sp[y].ls = x, pre = y;
    sp[y].pre = fa;
    if(root)
        root = false, sp[y].root = true;
    else
        sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y;
    update();    
}
void Splay::zag(int x)
{
    int y = ls, fa = pre;
    ls = sp[y].rs, sp[ls].pre = x;
    sp[y].rs = x, pre = y;
    sp[y].pre = fa;
    if(root)
        root = false, sp[y].root = true;
    else
        sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y;
    update();
}
void Splay::splay(int x)
{
    int y, z;
    for(pushdown(); !root;)
    {
        y = pre;
        if(sp[y].root) sp[y].pushdown(), pushdown(), sp[y].rs == x ? sp[y].zig(y) : sp[y].zag(y);
        else
        {
            z = sp[y].pre, sp[z].pushdown(), sp[y].pushdown(), pushdown();
            if(sp[z].rs == y)
            {
                if(sp[y].rs == x) sp[z].zig(z), sp[y].zig(y);
                else sp[y].zag(y), sp[z].zig(z);
            }
            else
            {
                if(sp[y].ls == x) sp[z].zag(z), sp[y].zag(y);
                else sp[y].zig(y), sp[z].zag(z);
            }    
        }
    }
    update();
}
void access(int x)
{
    int fx;
    for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre)
    {
        sp[fx].splay(fx);
        sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false;
        sp[fx].update();
    }    
}
void cut(int x)
{
    access(x), sp[x].splay(x);
    sp[sp[x].ls].root = true, sp[sp[x].ls].pre = 0, sp[x].ls = 0;
    sp[x].update();
}
void join(int x, int y, int c)
{
    access(y), sp[y].splay(y);
    sp[x].splay(x);
    if(sp[y].pre != 0) return ;
    cut(x);
    sp[x].pre = y, sp[x].col = c;
    sp[x].update();
}
void init()
{
    int i, x;
    for(i = 1; i <= N; i ++)
    {
        scanf("%d", &x);
        sp[i].init(), sp[i].pre = x;
    }
    for(i = 1; i <= N; i ++)
    {
        scanf("%d", &x);
        sp[i].col = x, sp[i].set = 1 << x, sp[i].size = 1;
    }
    sp[0].init();
}
void paint(int x, int y, int c)
{
    int fx;
    access(y), sp[y].splay(y);
    for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre)
    {
        sp[fx].splay(fx);
        if(sp[fx].pre == 0)
        {
            if(sp[y].pre != 0 || fx == y) same(sp[fx].rs, c), same(x, c);
        }
        sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false;
        sp[fx].update();    
    }    
}
void count(int x, int y)
{
    int fx;
    access(y), sp[y].splay(y);
    for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre)
    {
        sp[fx].splay(fx);
        if(sp[fx].pre == 0)
        {
            if(sp[y].pre == 0 && fx != y) printf("0 0\n");
            else
            {
                int cnt = 0, t = sp[sp[fx].rs].set | sp[x].set;
                for(int i = 1; i <= 30; i ++) if(t & 1 << i) ++ cnt;
                printf("%d %d\n", sp[sp[fx].rs].size + sp[x].size, cnt);
            }
        }
        sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false;
        sp[fx].update();    
    }    
}
void solve()
{
    int i, op, x, y, c;
    for(i = 0; i < M; i ++)
    {
        scanf("%d", &op);
        if(op == 1)
        {
            scanf("%d%d%d", &x, &y, &c);
            if(x == y) continue;
            join(x, y, c);
        }    
        else if(op == 2)
        {
            scanf("%d%d%d", &x, &y, &c);
            if(x == y) continue;
            paint(x, y, c);    
        }
        else
        {
            scanf("%d%d", &x, &y);
            if(x == y) printf("0 0\n");
            else count(x, y);    
        }
    }
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();    
    }
    return 0;    
}

转载于:https://www.cnblogs.com/staginner/archive/2012/08/23/2652982.html