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2243: [SDOI2011]染色树链剖分

对于线段树的操作,维护左端值,维护右端值,维护种类数,更新的时候,如果左儿子的右端==右儿子的左端,种类数减一,剩下的就是细节了。

#include<iostream>#include<cstdio>#include<cstring>#include<map>#include<vector>#include<stdlib.h>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn = 222222;int deep[maxn];struct Node{    int next; int to;}e[maxn * 2];int len, z; int head[maxn];int top[maxn], son[maxn], father[maxn], size[maxn], pos[maxn], val[maxn], vis[maxn];int color[maxn << 2], sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];void add(int from, int to){    e[len].to = to;    e[len].next = head[from];    head[from] = len++;}void init(int x){    size[x] = 1; son[x] = 0;    for (int i = head[x]; i != -1; i = e[i].next){        int cc = e[i].to;        if (cc == father[x]) continue;        father[cc] = x; deep[cc] = deep[x] + 1;        init(cc);        size[x] += size[cc];        if (size[son[x]] < size[cc]) son[x] = cc;    }}void dfs(int x, int tp){    pos[x] = ++z; vis[z] = x; top[x] = tp;    if (son[x]) dfs(son[x], tp);    for (int i = head[x]; i != -1; i = e[i].next){        int cc = e[i].to;        if (cc == son[x] || cc == father[x]) continue;        dfs(cc, cc);    }}void up(int rt){    lsum[rt] = lsum[rt << 1]; rsum[rt] = rsum[rt << 1 | 1];    if (rsum[rt << 1] == lsum[rt << 1 | 1]) sum[rt] = sum[rt << 1] + sum[rt << 1 | 1] - 1;    else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l, int r, int rt){    color[rt] = -1;    if (l == r){        color[rt] = lsum[rt] = rsum[rt] = val[vis[l]];        sum[rt] = 1;        return;    }    int mid = (l + r) >> 1;    build(lson);    build(rson);    up(rt);}void down(int rt){    if (~color[rt]){        sum[rt << 1] = sum[rt << 1 | 1] = 1;        lsum[rt << 1] = color[rt]; rsum[rt << 1] = color[rt];        color[rt << 1] = color[rt << 1 | 1] = color[rt];        lsum[rt << 1 | 1] = color[rt]; rsum[rt << 1 | 1] = color[rt];        color[rt] = -1;    }}void update(int L, int R, int ans, int l, int r, int rt){    if (L <= l&&r <= R){        color[rt] = ans; sum[rt] = 1;        lsum[rt] = ans; rsum[rt] = ans; return;    }    int mid = (l + r) >> 1;    down(rt);    if (L <= mid) update(L, R, ans, lson);    if (R > mid) update(L, R, ans, rson);    up(rt);}int ask(int L, int R, int l, int r, int rt){    if (L <= l&&r <= R) return sum[rt];    int ans = 0;    int mid = (l + r) >> 1;    down(rt);    if (L <= mid) ans += ask(L, R, lson);    if (R > mid) ans += ask(L, R, rson);    if (L <= mid&&R > mid&&rsum[rt << 1] == lsum[rt << 1 | 1]) ans--;    return ans;}void gao(int x, int y, int ans){    int fx = top[x]; int fy = top[y];    while (fx != fy){        if (deep[fx] < deep[fy]){            swap(x, y); swap(fx, fy);        }        update(pos[fx], pos[x], ans, 1, z, 1);        x = father[fx]; fx = top[x];    }    if (deep[x]>deep[y]) swap(x, y);    update(pos[x], pos[y], ans, 1, z, 1);}int ask1(int key, int l, int r, int rt){    if (l == r) return color[rt];    down(rt);    int mid = (l + r) >> 1;    if (key <= mid) return ask1(key, lson);    else return ask1(key, rson);}int gao1(int x, int y){    int ans = 0;    int fx = top[x]; int fy = top[y];    while (fx != fy){        if (deep[fx] < deep[fy]){            swap(x, y); swap(fx, fy);        }        ans += ask(pos[fx], pos[x], 1, z, 1);        x = father[fx];//手戳 ,写成了father[x],看了老半天        int t = ask1(pos[x], 1, z, 1);        int t1 = ask1(pos[fx], 1, z, 1);        if (t == t1) ans--;        fx = top[x];    }    if (deep[x]>deep[y]) swap(x, y);    ans += ask(pos[x], pos[y], 1, z, 1);//pos[x] 顺手写成了pos[son[x]],看了好久    return ans;}int main(){    char str[100];    int n, m;    int a, b, c;    while (cin >> n >> m){        z = len = 0;        memset(head, -1, sizeof(head));        for (int i = 1; i <= n; i++){            scanf("%d", &val[i]);        }        for (int i = 1; i <= n - 1; i++){            scanf("%d%d", &a, &b);            add(a, b); add(b, a);        }        deep[1] = 1;        init(1); dfs(1, 1); build(1, z, 1);        for (int i = 0; i < m; i++){            scanf("%s", str);            if (str[0] == C){                scanf("%d%d%d", &a, &b, &c);                gao(a, b, c);            }            else{                scanf("%d%d", &a, &b);                int t = gao1(a, b);                printf("%d\n", t);                //printf("%d %d %d %d jijiji\n",ask1(pos[1],1,z,1),ask1(pos[2],1,z,1),ask1(pos[3],1,z,1),ask1(pos[5],1,z,1));            }        }    }    return 0;}

 

2243: [SDOI2011]染色树链剖分