P2161 [SHOI2009]会场预约 - 线段树染色

Posted 2021-01-12 zolrk

是真的染色，把不同预约看做不同颜色，现在问题就是一个区间内不同颜色的数量，这个分块线段树都能做吧（不考虑复杂度用莫队也行）
注意，线段树的最大边界必须是定值，不能随输入改变（一开始懒得离线动态更新右端点然后节点的编号就串了）
注意数组大小，因为same和tag数组都是针对线段树节点设置的，所以其数组大小也要开4倍

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 100000 + 10;
typedef long long ll;
int n,m,eff,rig,fail,tag[MAXN * 4],same[MAXN * 4],qx,qy,num,del[MAXN * 4],era;
char cmd[5];
struct segment_tree{
    int val, add;
}tr[MAXN * 4];

struct opts{
    int cmd, st, ed;
}opt[MAXN * 2];

void build(int w, int l, int r) {
    same[w] = 1, tag[w] = 0;
    if(l == r) return;
    int mid = l + r >> 1;
    build(w*2, l, mid);
    build(w*2+1, mid+1, r);
}

void down(int w, int l, int r) {
    same[w] = 0;
    if(!tag[w]) return;
    int le = w * 2, ri = w * 2 + 1; 
    tag[le] = tag[ri] = tag[w];
    tag[w] = 0;
}

void find(int w, int l, int r) {
    if(same[w] == 1) {
        if(!del[tag[w]] && tag[w]) 
            eff--, era++;
        del[tag[w]] = 1;
        tag[w] = num;
        return;
    }
    int le = w * 2, ri = w * 2 + 1; 
    int mid = l + r >> 1;
    find(le, l, mid), find(ri, mid+1, r);
    tag[w] = num, same[w] = 1;
}


void change(int w, int l, int r) {
    if(qx <= l && r <= qy) {
        find(w, l, r);
        return;
    }
    down(w, l, r);
    int le = w * 2, ri = w * 2 + 1; 
    int mid = l + r >> 1;
    if(qx <= mid)
        change(le, l, mid);
    if(qy > mid)
        change(ri, mid+1, r);
}

int main() {
    scanf("%d", &n);
    char cmd[5];
    for(int i=1; i<=n; i++) {
        scanf("%s", cmd);
        if(cmd[0] == 'A') {
            int sta, ed;
            scanf("%d%d", &sta, &ed);
            rig = max(rig, ed);
            opt[i].cmd = 1;
            opt[i].st = sta;
            opt[i].ed = ed;
        } else {
            opt[i].cmd = 2;
        }
    }
    build(1, 1, rig);
    for(int i=1; i<=n; i++) {
        if(opt[i].cmd == 1) {
            int sta = opt[i].st, ed = opt[i].ed;
            eff++;          
            num++;
            era = 0;
            qx = sta, qy = ed;
            change(1, 1, rig);
            printf("%d
", era);
        } else {
            printf("%d
", eff);
        }
    } 
    return 0;
}