【HDU 5372】Segment Game（树状数组）

题解：

对于新插入的线段，查询有多少个线段左端点大于等于该线段的左端点。 再查询有多少个线段的右端点大于该线段右端点， 两者之差就是答案。

这里注意两个问题，一个是离散化，第二个这道题时间卡的可能比较严，线段树貌似会超时~

好久没写离散化了。。。生疏了

#include
#include
#include
using namespace std;
const int maxn = 1000005;
int n;
//----------------------------------------------------------
int Hash[maxn * 2];
int Hash_num;
int HASH(int u){
    return lower_bound(Hash + 1,Hash + Hash_num,u) - Hash;
}
void Hash_debug(){
    for(int i = 1; i < Hash_num; i++)
        printf("%d ",Hash[i]);
    puts("");
}
//-----------------------------------------------------------
int C[2][maxn];
int lowbit(int x){
    return  x & (-x);
}
int sum(int op,int x){
    int ret = 0;
    while(x > 0){
        ret += C[op][x];
        x -= lowbit(x);
    }
    return ret;
}
void add(int op,int x,int value){
    while(x <= Hash_num){
        C[op][x] += value;
        x += lowbit(x);
    }
}
//-----------------------------------------------------------
struct Input{
    int op,id,lv;
}input[maxn];
int pid[maxn];
//------------------------------------------------------------
int main(){
    int Case = 1;
    while(scanf("%d",&n) != EOF){
        int cnt = 1;
        Hash_num = 1;
        Hash[Hash_num++] = 0;
        memset(C,0,sizeof(C));
        for(int i = 0; i < n; i++){
            scanf("%d%d",&input[i].op,&input[i].lv);
            if(input[i].op == 0){
                input[i].lv ++;
                Hash[Hash_num++] = input[i].lv;
                Hash[Hash_num++] = input[i].lv + cnt;
                pid[cnt] = i;
                input[i].id = cnt ++;
            }
        }
        sort(Hash + 1,Hash + Hash_num);
        Hash_num = unique(Hash + 1,Hash + Hash_num) - Hash;
        //Hash_debug();
        printf("Case #%d:\n",Case++);
        for(int i = 0; i < n; i++){
            if(input[i].op == 0){
                int sum1 = sum(0,Hash_num);
                int sum2 = sum(1,Hash_num);
                int lv   = HASH(input[i].lv);
                int rv   = HASH(input[i].lv + input[i].id);
                int sum11 = sum1 - sum(0,lv - 1);
                int sum22 = sum2 - sum(1,rv);
                printf("%d\n",abs(sum11 - sum22));
                add(0,lv,1);
                add(1,rv,1);
            }
            else{
                int id = pid[input[i].lv];
                int lv = HASH(input[id].lv);
                int rv = HASH(input[id].lv + input[id].id);
                add(0,lv,-1);
                add(1,rv,-1);
            }
        }
    }
    return 0;
}