hdu 4419 线段树 扫描线 离散化 矩形面积

//离散化 + 扫描线 + 线段树
//这个线段树跟平常不太一样的地方在于记录了区间两个信息，len[i]表示颜色为i的被覆盖的长度为len[i]， num[i]表示颜色i 『完全』覆盖了该区间几层。len[i]起传递儿子与父亲的关系，而num[i]不起传递作用，只是单纯的表示被覆盖的区间。
//然后就是pushUp函数，必须在update到底层后即更新num[]和len，然后把len传上去。
//离散化后由于求的是面积，所以我是把每条长度为1的线段当做一个点， 即把左端点表示此段长度。而不是把点当成点。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
#define lson l, m, rt<<1
#define rson m + 1, r, rt<<1|1
typedef long long ll;
const int maxn = 2e4 + 5;

struct Seg{
    char c;
    int y, s, t, tag;
}ss[maxn];
bool cmp(Seg a, Seg b){
    return a.y < b.y;
}
int san[maxn], tot;
int num[maxn << 2][5],len[maxn << 2][8];
ll ans[8];
void pushUp(int l, int r, int rt){
    int state = (num[rt][1] > 0 ? 1 : 0) | (num[rt][2] > 0 ? 2 : 0) | (num[rt][4] > 0 ? 4 : 0);
    memset(len[rt], 0, sizeof(len[rt]));
    if (state){
        len[rt][state] = san[r] - san[l - 1];
        for (int i = 1; i < 8; ++i){
            if (state != (state|i)){
                int tmp = len[rt<<1][i] + len[rt<<1|1][i];
                len[rt][state|i] += tmp;
                len[rt][state] -= tmp;
            }
        }
    }
    else if (l != r){
        for (int i = 1; i < 8; ++i) len[rt][i] = len[rt<<1][i] + len[rt<<1|1][i];
    }
}
int getC(char c){
    if (c == 'R') return 1;
    if (c == 'G') return 2;
    return 4;
}
void update(int L, int R, char c, int tag, int l, int r, int rt){
    if (L <= l && R >= r){
        int cc = getC(c);
        num[rt][cc] += tag;
        //注意
        pushUp(l, r, rt);
        return ;
    }
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, c, tag, lson);
    if (R > m) update(L, R, c, tag, rson);
    pushUp(l, r, rt);
}
int T,  n;
int main(){
    int tcas = 0;
    int x1, x2, y1, y2;
    char s[3];
    scanf("%d", &T);
    while (T--){
        scanf("%d", &n);
        tot = 0;
        for (int i = 1; i <= n; ++i){
            scanf("%s%d%d%d%d", s,&x1, &y1, &x2, &y2); 
            ss[i].c = s[0]; ss[i].y = y1; ss[i].s = x1; ss[i].t = x2, ss[i].tag= 1;
            ss[i + n].c = s[0]; ss[i + n].y = y2; ss[i + n].s = x1; ss[i + n].t = x2, ss[i + n].tag = -1;
            san[tot++] = x1 ; san[tot++] = x2;
        }
        n = n * 2;

        sort(san, san + tot);
        tot = unique(san, san + tot) - san;
        sort(ss + 1, ss + n + 1, cmp);
        ss[0].y = ss[1].y;
        
        memset(num, 0, sizeof(num));
        memset(len, 0, sizeof(len));
        memset(ans, 0, sizeof(ans));
        for (int i = 1; i <= n; ++i){
            int l = lower_bound(san, san + tot, ss[i].s) - san + 1;
            int r = lower_bound(san, san + tot, ss[i].t) - san;
            /*cout << " l = " << l << " r = " << r ;
            cout << " tag = " << ss[i].tag << " c = "<< ss[i].c << endl;
            for (int j = 1; j < 8; ++j) cout << len[1][j] << " ";
            cout << endl;
            cout << endl;
            */
            if (ss[i].y != ss[i - 1].y){
                for (int j = 1; j < 8; ++j){
                    ans[j] += (ll)(ss[i].y - ss[i - 1].y) * (ll)len[1][j];
                }
            }
            update(l, r, ss[i].c, ss[i].tag, 1, tot - 1, 1);
        }
        printf("Case %d:
", ++tcas);
        swap(ans[3], ans[4]);
        for (int i = 1; i < 8; ++i)
            printf("%lld
", ans[i]);
    }
    return 0;
}