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bzoj 2243: SDOI2011 染色

时间:2014-08-31 17:12:41      阅读:196      评论:0      收藏:0      [点我收藏+]

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    最近总是在做树链剖分的题(觉得有必要学一下倍增算法=_=)。这题也是一个树链剖分。维护和找答案的时候注意区间左右端点的颜色就OK了……

    上代码:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#define N 100010
#define inf 0x7f7f7f7f
using namespace std;

struct sss
{
    int color, lc, rc;
    int num, push;
}t[N*4];
int n, m, color[N], nowplace = 0;
int fa[N], deep[N], w[N], top[N], son[N] = {0}, siz[N] = {0};
int p[N] = {0}, v[N*2], next[N*2], bnum = 0;

void build_tree(int now, int l, int r)
{
    t[now].push = 0; t[now].num = 0;
    t[now].color = t[now].lc = t[now].rc = 0;
    if (l == r) return;
    int mid = (l+r) /2 ;
    build_tree(now*2, l, mid); build_tree(now*2+1, mid+1, r);
}

void update(int now)
{
    t[now].num = t[now*2].num + t[now*2+1].num;
    if (t[now*2].rc == t[now*2+1].lc) t[now].num--;
    t[now].lc = t[now*2].lc; t[now].rc = t[now*2+1].rc;
    if (t[now*2].color && t[now*2+1].color == t[now*2].color)
        t[now].color = t[now*2].color;
    else t[now].color = 0;
}

void downdate(int now)
{
    if (!t[now].push) return;
    t[now*2].push = t[now*2+1].push = 1; t[now].push = 0;
    t[now*2].color = t[now*2+1].color = t[now].color;
    t[now*2].lc = t[now*2].rc = t[now*2+1].rc = t[now*2+1].lc = t[now].color;
    t[now*2].num = t[now*2+1].num = 1;
}

int findcolor(int now, int l, int r, int place)
{
    if (l == r) return t[now].color;
    int mid = (l+r) / 2;
    downdate(now);
    if (place <= mid) return findcolor(now*2, l, mid, place);
    else return findcolor(now*2+1, mid+1, r, place);
}

void make_tree(int now, int l, int r)
{
    if (l == r)
    {
        t[now].color = t[now].lc = t[now].rc = 0;
        t[now].num = 0; t[now].push = 0; return;
    }
    int mid = (l+r) / 2;
    make_tree(now*2, l, mid); make_tree(now*2+1, mid+1, r);
}

void tchange(int now, int l, int r, int cl, int cr, int cnum)
{
    if (cl <= l && r <= cr)
    {
        t[now].color = cnum; t[now].lc = t[now].rc = cnum;
        t[now].num = 1; t[now].push = 1; return;
    }
    int mid = (l+r) / 2; downdate(now);
    if (cl <= mid) tchange(now*2, l, mid, cl, cr, cnum);
    if (cr > mid) tchange(now*2+1, mid+1, r, cl, cr, cnum);
    update(now);
}

int task(int now, int l, int r, int al, int ar)
{
    if (al <= l && r <= ar) return t[now].num;
    int mid = (l+r) / 2, ans = 0; downdate(now);
    if (al <= mid) ans += task(now*2, l, mid, al, ar);
    if (ar > mid) ans += task(now*2+1, mid+1, r, al, ar);
    if (al <= mid && ar > mid)
        if (findcolor(1, 1, n, mid) == findcolor(1, 1, n, mid+1))
            ans--;
    return ans;
}

void addbian(int x, int y)
{
    bnum++; next[bnum] = p[x]; p[x] = bnum; v[bnum] = y;
    bnum++; next[bnum] = p[y]; p[y] = bnum; v[bnum] = x;
}

void dfs_1(int now, int fat, int nowdeep)
{
    int k = p[now]; fa[now] = fat; deep[now] = nowdeep;
    int maxson = 0; siz[now] = 1; 
    while (k)
    {
        if (v[k] != fat)
        {
            dfs_1(v[k], now, nowdeep+1);
            siz[now] += siz[v[k]];
            if (siz[v[k]] > maxson)
            {
                maxson = siz[v[k]];
                son[now] = v[k];
            }
        }
        k = next[k];
    }
}

void dfs_2(int now, int fat, int nowtop)
{
    int k = p[now]; w[now] = ++nowplace; top[now] = nowtop;
    tchange(1, 1, n, w[now], w[now], color[now]);
    if (son[now]) dfs_2(son[now], now, nowtop);
    while (k)
    {
        if (v[k] != fat && v[k] != son[now])
            dfs_2(v[k], now, v[k]);
        k = next[k];
    }
}

void change(int u, int v, int changenum)
{
    int f1 = top[u], f2 = top[v];
    if (deep[f1] < deep[f2]) { swap(f1, f2); swap(u, v); }
    if (f1 == f2)
    {
        if (u == v) tchange(1, 1, n, w[u], w[u], changenum);
        else tchange(1, 1, n, min(w[u], w[v]), max(w[u], w[v]), changenum);
    }
    else
    {
        tchange(1, 1, n, w[f1], w[u], changenum);
        change(fa[f1], v, changenum);
    }
}

int ask(int u, int v)
{
    int f1 = top[u], f2 = top[v], ans = 0;
    if (deep[f1] < deep[f2]) { swap(f1, f2); swap(u, v); }
    if (f1 == f2)
    {
        if (u == v) return ans+1;
        else return task(1, 1, n, min(w[u], w[v]), max(w[u], w[v]));
    }
    else
    {
        ans += task(1, 1, n, w[f1], w[u]);
        int x = findcolor(1, 1, n, w[f1]), y = findcolor(1, 1, n, w[fa[f1]]);
        if (x == y) ans --;
        ans += ask(fa[f1], v);
        return ans;
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    build_tree(1, 1, n);
    for (int i = 1; i <= n; ++i)
        scanf("%d", &color[i]);
    for (int i = 1; i < n; ++i)
    {
        int x, y; scanf("%d%d", &x, &y);
        addbian(x, y);
    }
    dfs_1(1, 0, 1);
    dfs_2(1, 0, 1);
    for (int i = 1; i <= m; ++i)
    {
        char s[2]; scanf("%s", s);
        if (s[0] == C)
        {
            int a, b, c; scanf("%d%d%d", &a, &b, &c);
            change(a, b, c);
        }
        else
        {
            int a, b; scanf("%d%d", &a, &b);
            printf("%d\n", ask(a, b));
        }
    }
    return 0;
}

 

bzoj 2243: SDOI2011 染色

标签:style   blog   color   os   io   ar   for   div   代码   

原文地址:http://www.cnblogs.com/handsomeJian/p/3947805.html

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