BZOJ 2243 染色

树链剖分+区间染色

因为是一颗树不是森林，所以应该用树剖就行，但是LCT好像也能写。。
直接用线段树维护树上的节点，注意pushdown还有询问的时候要考虑区间相交的地方，也就是左孩子右边和有孩子的左边，如果两个颜色相同就-1
树上询问的时候也是一样，跨越轻链的时候也要看一下相接的地方颜色是不是一样。。w

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 100005;
int n, m, cnt, dfn, head[N], size[N], depth[N], p[N], son[N], top[N], val[N], w[N], id[N];
int tree[N<<2], lc[N<<2], rc[N<<2], c[N<<2];
struct Edge { int v, next; } edge[N<<1];

void addEdge(int a, int b){
    edge[cnt].v = b, edge[cnt].next = head[a], head[a] = cnt ++;
}

void dfs1(int s, int fa){
    depth[s] = depth[fa] + 1;
    p[s] = fa;
    size[s] = 1;
    int child = -1;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == fa) continue;
        dfs1(u, s);
        size[s] += size[u];
        if(size[u] > child) child = size[u], son[s] = u;
    }
}

void dfs2(int s, int tp){
    id[s] = ++dfn;
    w[id[s]] = val[s];
    top[s] = tp;
    if(son[s] != -1) dfs2(son[s], tp);
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == p[s] || u == son[s]) continue;
        dfs2(u, u);
    }
}

void push_up(int rt){
    int l = rt << 1, r = rt << 1 | 1;
    tree[rt] = tree[l] + tree[r];
    lc[rt] = lc[l], rc[rt] = rc[r];
    if(rc[l] == lc[r]) tree[rt] --;
}

void push_down(int rt){
    //printf("c[%d] = %d\n", rt, c[rt]);
    if(c[rt] != -1){
        int ls = rt << 1, rs = rt << 1 | 1;
        c[ls] = c[rs] = c[rt];
        tree[ls] = tree[rs] = 1;
        lc[ls] = rc[ls] = lc[rs] = rc[rs] = c[rt];
        c[rt] = -1;
    }
}

void buildTree(int rt, int l, int r){
    if(l == r){
        tree[rt] = 1, c[rt] = -1, lc[rt] = rc[rt] = w[l];
        return;
    }
    int mid = (l + r) >> 1;
    buildTree(rt << 1, l, mid);
    buildTree(rt << 1 | 1, mid + 1, r);
    push_up(rt);
}

void modify(int rt, int l, int r, int ml ,int mr, int tot){
    if(l == ml && r == mr){
        c[rt] = tot, tree[rt] = 1, lc[rt] = rc[rt] = tot;
        return;
    }
    push_down(rt);
    int mid = (l + r) >> 1;
    if(ml > mid) modify(rt << 1 | 1, mid + 1, r, ml, mr, tot);
    else if(mr <= mid) modify(rt << 1, l, mid, ml, mr, tot);
    else modify(rt << 1, l, mid, ml, mid, tot), modify(rt << 1 | 1, mid + 1, r, mid + 1, mr, tot);
    push_up(rt);
}

int query(int rt, int l, int r, int ql, int qr){
    if(l == ql && r == qr){
        return tree[rt];
    }
    push_down(rt);
    int mid = (l + r) >> 1;
    if(ql > mid) return query(rt << 1 | 1, mid + 1, r, ql, qr);
    else if(qr <= mid) return query(rt << 1, l, mid, ql, qr);
    else{
        int ls = query(rt << 1, l, mid, ql, mid);
        int rs = query(rt << 1 | 1, mid + 1, r, mid + 1, qr);
        return rc[rt << 1] == lc[rt << 1 | 1] ? ls + rs - 1 : ls + rs;
    }
}

int color(int rt, int l, int r, int index){
    if(l == r) return lc[rt];
    push_down(rt);
    int mid = (l + r) >> 1;
    if(index > mid) return color(rt << 1 | 1, mid + 1, r, index);
    return color(rt << 1, l, mid, index);
}

void treeModify(int x, int y, int tot){
    while(top[x] != top[y]){
        if(depth[top[x]] < depth[top[y]]) swap(x, y);
        modify(1, 1, n, id[top[x]], id[x], tot);
        x = p[top[x]];
    }
    if(depth[x] > depth[y]) swap(x, y);
    modify(1, 1, n, id[x], id[y], tot);
}

int treeQuery(int x, int y){
    int ret = 0;
    while(top[x] != top[y]){
        if(depth[top[x]] < depth[top[y]]) swap(x, y);
        ret += query(1, 1, n, id[top[x]], id[x]);
        if(color(1, 1, n, id[top[x]]) == color(1, 1, n, id[p[top[x]]])) ret --;
        x = p[top[x]];
    }
    if(depth[x] > depth[y]) swap(x, y);
    ret += query(1, 1, n, id[x], id[y]);
    return ret;
}

int main(){

    full(head, -1), full(son, -1), full(c, -1);
    n = read(), m = read();
    for(int i = 1; i <= n; i ++) val[i] = read();
    for(int i = 0; i < n - 1; i ++){
        int u = read(), v = read();
        addEdge(u, v), addEdge(v, u);
    }
    dfs1(1, 0), dfs2(1, 1);
    buildTree(1, 1, n);
    while(m --){
        char opt[10]; scanf("%s", opt);
        int a = read(), b = read();
        if(opt[0] == 'C'){
            int c = read();
            treeModify(a, b, c);
        }
        else if(opt[0] == 'Q'){
            printf("%d\n", treeQuery(a, b));
        }
    }
    return 0;
}

BZOJ 2243 染色

原文：https://www.cnblogs.com/onionQAQ/p/10712150.html