loj10141. 「一本通 4.5 练习 3」染色

思路：

　　利用树链剖分转化为线段树问题，考虑线段上经这两种操作后的sum求值方法，并着重考虑树链合并的情况即可。线段树为了打cov,要注意将所有的颜色统一加一。

#include<cstdio>
#include<iostream>
using namespace std;
const int maxn = 100010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘)    ch = getchar();
    while(ch >= ‘0‘ && ch <= ‘9‘)    x = 10 * x + ch - 48, ch = getchar();
}
int n, q;
int head[maxn];
int go[maxn << 1];
int nxt[maxn << 1];

int val[maxn];
int size[maxn];
int son[maxn];
int deep[maxn];
int f[maxn];
int top[maxn];
int seg[maxn];
int rev[maxn];
int end[maxn];

int sum[maxn << 2];
int cov[maxn << 2];
inline void init(){
    qread(n);
    qread(q);
    for(int i=1; i<=n; ++i){
        qread(val[i]);
        ++val[i];
    }
    for(int i=1; i<n; ++i){
        int x, y;
        qread(x), qread(y);
        go[i] = y;
        nxt[i] = head[x];
        head[x] = i;
        go[i + n] = x;
        nxt[i + n] = head[y];
        head[y] = i + n;
    }
    deep[1] = top[1] = seg[0] = seg[1] = rev[1] = 1;
}
void dfs1(int x){
    size[x] = 1;
    for(register int i = head[x]; i; i = nxt[i])
        if(!deep[go[i]]){
            f[go[i]] = x;
            deep[go[i]] = deep[x] + 1;
            dfs1(go[i]);
            size[x] += size[go[i]];
            if(size[go[i]] > size[son[x]])
                son[x] = go[i];
        }
}
void dfs2(int x){
    end[x] = x;
    if(son[x]){
        top[son[x]] = top[x];
        seg[son[x]] = ++seg[0];
        rev[seg[0]] = son[x];
        dfs2(son[x]);
        end[x] = end[son[x]];
    }
    for(register int i = head[x]; i; i = nxt[i])
        if(!top[go[i]]){
            top[go[i]] = go[i];
            seg[go[i]] = ++seg[0];
            rev[seg[0]] = go[i];
            dfs2(go[i]);
            end[x] = end[go[i]];
        }
}
void build(int l, int r, int x){
    if(l == r){
        sum[x] = 1;
        return ;
    }
    int mid = (l + r) >> 1;
    build(l, mid, x << 1);
    build(mid + 1, r, x << 1 | 1);
    sum[x] = sum[x << 1] + sum[x << 1 | 1] - (val[rev[mid]] == val[rev[mid + 1]]);
}
inline void pushdown(int x){
    cov[x << 1] = cov[x];
    cov[x << 1 | 1] = cov[x];
    sum[x << 1] = 1;
    sum[x << 1 | 1] = 1;
    cov[x] = 0;
}
int visit(int l, int r, int P, int x){
    if(l == r)
        return cov[x] ? cov[x] : val[rev[l]];
    if(cov[x])    pushdown(x);
    int mid = (l + r) >> 1;
    if(mid >= P)    return visit(l, mid, P, x << 1);
    else            return visit(mid + 1, r, P, x << 1 | 1);
}
void change(int l, int r, int L, int R, int v, int x){
    if(L <= l && R >= r){
        cov[x] = v;
        sum[x] = 1;
        return ;
    }
    if(cov[x])    pushdown(x);
    int mid = (l + r) >> 1;
    if(mid >= L)    change(l, mid, L, R, v, x << 1);
    if(mid < R)        change(mid + 1, r, L, R, v, x << 1 | 1);
    sum[x] = sum[x << 1] + sum[x << 1 | 1] - (visit(1, n, mid, 1) == visit(1, n, mid + 1, 1));
}
int ask(int l, int r, int L, int R, int x){
    if(L <= l && R >= r)
        return sum[x];
    if(cov[x])    pushdown(x);
    int mid = (l + r) >> 1, cnt = 0, ans = 0;
    if(mid >= L)    ++cnt, ans += ask(l, mid, L, R, x << 1);
    if(mid < R)        ++cnt, ans += ask(mid + 1, r, L, R, x << 1 | 1);
    if(cnt == 2)    ans -= (visit(1, n, mid, 1) == visit(1, n, mid + 1, 1));
    return ans;
}
int treeask(int x, int y){
    int fx = top[x], fy = top[y], ans = 0;
    while(fx != fy){
        if(deep[fx] < deep[fy])    swap(x, y), swap(fx, fy);
        ans += ask(1, n, seg[fx], seg[x], 1);
        if(f[fx] != 0 && visit(1, n, seg[fx], 1) == visit(1, n, seg[f[fx]], 1))    ans--;
        x = f[fx];
        fx = top[x];
    }
    if(deep[x] > deep[y])    swap(x, y);
    ans += ask(1, n, seg[x], seg[y], 1);
    return ans;
}
void treechange(int x, int y, int z){
    int fx = top[x], fy = top[y];
    while(fx != fy){
        if(deep[fx] < deep[fy])    swap(x, y), swap(fx, fy);
        change(1, n, seg[fx], seg[x], z, 1);
        x = f[fx];
        fx = top[x];
    }
    if(deep[x] > deep[y])    swap(x, y);
    change(1, n, seg[x], seg[y], z, 1);
}

int main(void)
{
    init();
    dfs1(1);
    dfs2(1);
    build(1, n, 1);
    while(q--){
        string op;
        cin >> op;
        if(op == "C") {
            int a, b, c;
            qread(a), qread(b), qread(c);
            ++c;
            treechange(a, b, c);
        }
        else{
            int a, b;
            qread(a), qread(b);
            printf("%d\n", treeask(a, b));
        }
    }
}

原文地址：https://www.cnblogs.com/junk-yao-blog/p/9485071.html