POJ2528线段树区间合并加离散化

POJ2528Mayor’s posters就只支持一种操作，一次性的给一段区间涂上颜色，且每次颜色均不一样。问最后一共可以看到多少种颜色。
关于线段树的更详细实现请参考线段树解决区间问题包括延迟操作以及离散化

/*
    在数轴上，一次给一个线段涂上颜色
    后面的颜色会覆盖前面的颜色
    问最后能看到多少个颜色

    显然是成段更新，线段树
    区间范围是1千万，需要离散化
    另外注意位置编号的含义
*/

#include 
#include 
#include 
#include 
using namespace std;

int const SIZE = 20010;

inline int lson(int x){return x<<1;}
inline int rson(int x){return lson(x)|1;}

int ST[SIZE<<2];//段数
int L[SIZE<<2],R[SIZE<<2];//左右边界的颜色
int Lazy[SIZE<<2];

struct _t{
    int cnt;
    int left;
    int right;
    _t(int aa=0,int bb=0,int cc=0):cnt(aa),left(bb),right(cc){}
};

inline void _pushUp(int t){
    ST[t] = ST[lson(t)] + ST[rson(t)] + (R[lson(t)]==L[rson(t)]?-1:0);
    L[t] = L[lson(t)];
    R[t] = R[rson(t)];
}

inline void _pushDown(int t){
    if ( Lazy[t] ){
        ST[lson(t)] = ST[rson(t)] = 1;
        Lazy[lson(t)] = L[lson(t)] = R[lson(t)] = Lazy[t];
        Lazy[rson(t)] = L[rson(t)] = R[rson(t)] = Lazy[t];
        Lazy[t] = 0;
    }
}

_t query(int t,int s,int e,int a,int b){
    if ( a <= s && e <= b ){
        return _t(ST[t],L[t],R[t]);
    }

    _pushDown(t);
    int mid = (s+e) >> 1;
    _t lans,rans;
    if ( a <= mid ) {
        lans = query(lson(t),s,mid,a,b);
        if ( b <= mid ) return lans;
    }
    if ( mid < b ){
        rans = query(rson(t),mid+1,e,a,b);
        if ( mid < a ) return rans;
    }
    return _t(lans.cnt+rans.cnt+(lans.right==rans.left?-1:0),lans.left,rans.right);
}

void modify(int t,int s,int e,int a,int b,int color){
    if ( a <= s && e <= b ){
        ST[t] = 1;
        Lazy[t] = L[t] = R[t] = color;
        return;
    }

    _pushDown(t);//?
    int mid = ( s + e ) >> 1;
    if ( a <= mid ) modify(lson(t),s,mid,a,b,color);
    if ( mid < b ) modify(rson(t),mid+1,e,a,b,color);
    _pushUp(t);
}

int getUnsigned(){
	char ch = getchar();
	while( ch < '0' || ch > '9' ) ch = getchar();

	int ret = (int)(ch-'0');
	while( '0' <= ( ch = getchar() ) && ch <= '9' ) ret = ret * 10 + (int)(ch-'0');
	return ret;
}

int N,NN;
int T[SIZE];
int A[SIZE],B[SIZE];

bool read(){
    N = getUnsigned();
    for(int i=0;i<N;++i) T[i+i] = A[i] = getUnsigned(), T[i+i+1] = B[i] = getUnsigned() + 1;

    sort(T,T+N+N);
    NN = unique(T,T+N+N) - T;

    fill(ST,ST+NN*4,0);
    fill(Lazy,Lazy+NN*4,0);
    fill(L,L+NN*4,0);
    fill(R,R+NN*4,0);

    return true;
}

int Flag[SIZE];
int main(){
    //freopen("1.txt","r",stdin);

    int nofkase = getUnsigned();
    while(nofkase--){
        read();

        //成段更新
        for(int i=0;i<N;++i){
            int a = lower_bound(T,T+NN,A[i]) - T + 1;
            int b = lower_bound(T,T+NN,B[i]) - T;//左闭右开，就不用加1了
            modify(1,1,NN,a,b,i+1);
        }

        //检查结果
        fill(Flag,Flag+N+1,0);
        int cnt = 0;
        for(int i=1;i<=NN;++i){
            _t ans = query(1,1,NN,1,i);

            if ( ans.cnt > cnt && ans.right != 0 ){
                cnt = ans.cnt;
                Flag[ans.right] = 1;
            }
        }

        printf("%d\n",accumulate(Flag+1,Flag+N+1,0));
    }
    return 0;
}