注意到一个很重要的性质：每次添加的线段必定比前面添加的线段长。

所以可以用  左端点>=当前线段左端点的线段数 - 右端点>当前线段右端点的线段数，就是答案。

Segment Game

3
0 0
0 3
0 1
5
0 1
0 0
1 1
0 1
0 0

Case #1:
0
0
0
Case #2:
0
1
0
2
Hint
For the second case in the sample:

At the first add operation,Lillian adds a segment [1,2] on the line.

At the second add operation,Lillian adds a segment [0,2] on the line.

At the delete operation,Lillian deletes a segment which added at the first add operation.

At the third add operation,Lillian adds a segment [1,4] on the line.

At the fourth add operation,Lillian adds a segment [0,4] on the line
 

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const int N = 202002 << 1;
/****************/
struct Tree
{
    int tree[N];
    int low(int x) { return x & -x; }
    void clear() { memset(tree,0, sizeof tree); }
    void add(int p, int x)
    {
        if (p<=0) return;
        for (int i=p;i<N;i+=low(i))
            tree[i]+=x;
    }
    int sum(int p)
    {
        int r = 0;
        for (int i=p;i>0;i-=low(i)) r += tree[i];
        return r;
    }
}TL, TR;
/****************/
int n;
struct data
{
    int op, id;
}a[N];
struct P
{
    int l, r;
}p[N];
int t[N<<1], tt;
int id[N];
int main()
{
    int ca = 1;
    while (scanf("%d",&n)==1) {
        int i = 0;
        for (int i=1;i<=n;i++) scanf("%d%d", &a[i].op, &a[i].id);
        tt = 0;
        memset(id, -1, sizeof id);
        for (int j=1;j<=n;j++) {
            int op = a[j].op;
            if (op==0) {
                ++i;
                p[i].l = a[j].id;
                p[i].r = a[j].id + i;

                id[j] = i;
                t[tt++] = p[i].l;
                t[tt++] = p[i].r;
            }
            else {
                ;
            }
        }
        int m = i;
        sort(t, t+tt);
        tt = unique(t, t+tt) - t;
        map<int, int> Lisan;
        for (int i=0;i<tt;i++) {
            Lisan[t[i]] = i + 1;
        }
        for (int i=1;i<=m;i++) {
            p[i].l = Lisan[p[i].l];
            p[i].r = Lisan[p[i].r];
        }
        TL.clear();
        TR.clear();
        printf("Case #%d:\n", ca++);
        for (int i=1;i<=n;i++) {
            if (a[i].op == 0) {
                int j = id[i];

                int ans = j - 1 - TL.sum(p[j].l - 1);
                ans -= j - 1 - TR.sum(p[j].r);
                printf("%d\n", ans);
                TL.add(p[j].l, 1);
                TR.add(p[j].r, 1);
            }
            if (a[i].op == 1) {
                int j = a[i].id;
                TL.add(p[j].l, -1);
                TR.add(p[j].r, -1);
            }
        }
    }
    return 0;
}

HDU 5372 Segment Game

原文：http://blog.csdn.net/oilover/article/details/47442035