CodeForces540E【树状数组+二分】

思路：
1:变换的做一次树状数组求逆序就好了
2:然后就是对于变换的求前面有多少个比他大，后面有多少个比他小
具体：
对于当前位置，前面有多少比他大，然后减去被占多少位置（二分就好）
对于后面位置，后面有多少比他小，然后减去被占多少位置（二分就好）
然后就是模拟了。。然后就过了~

#include 
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
#define lson num<<1,Left,Mid
#define rson num<<1|1, Mid+1, Right
typedef long long LL;

const int Maxn = 2e5 + 10;
LL c[Maxn];
inline int lowbit(int x){return x & -x;}
LL Sum(int x){
    LL ans = 0;
    while(x){
        ans = ans + c[x];
        x = x - lowbit(x);
    }
    return ans;
}
void add(int x){
    while(x < Maxn){
        ++c[x];
        x = x + lowbit(x);
    }
}
vector<int>xs, p;
int getid(int x){return lower_bound(xs.begin(), xs.end(), x) - xs.begin() + 1;}
struct asd{
    int id, w;
}node[Maxn];
bool cmp(asd x, asd y){
    return x.id > y.id;
}
map<int, int>mp;
int n;
int p1, p2, w1, w2;
void solve(){
    if(mp.find(p1) == mp.end()) w1 = p1;
    else w1 = mp[p1];
    if(mp.find(p2) == mp.end()) w2 = p2;
    else w2 = mp[p2];
    mp[p1] = w2;
    mp[p2] = w1;
}

int main(){
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&p1,&p2);
        solve();
        xs.push_back(mp[p1]);
        xs.push_back(mp[p2]);
        p.push_back(p1);
        p.push_back(p2);
    }
    sort(p.begin(), p.end()), p.erase(unique(p.begin(), p.end()), p.end());
    sort(xs.begin(), xs.end()), xs.erase(unique(xs.begin(), xs.end()), xs.end());
    int num = 0;
    int sz = p.size();
    for(int i=0;i0;
    for(int i=0;i1);
    }
//    printf("%d\n" , ans);

    int p1, p2, it1, it2, Left, Right;
    LL pos, w, len, ext;
    for(int i=0;iif(pos > w){
            len = pos - w - 1;
            p1 = w + 1;
            p2 = pos - 1;
            Left = 0, Right = p.size()-1;
            while(Left < Right){
                int Mid = Left + (Right - Left) / 2;
                if(p[Mid] >= p1) Right = Mid;
                else Left = Mid + 1;
            }
            it1 = Left;
            Left = 0, Right = p.size()-1;
            while(Left < Right){
                int Mid = Left + (Right - Left) / 2;
                if(p[Mid] > p2) Right = Mid;
                else Left = Mid + 1;
            }
            it2 = Left;
            if(p[it1] >= p1 && p[it2] > p2){
                ext = it2 - it1;
//                printf("%I64d\n", ext);
                ans = ans + len - ext;
            }
        }
        if(w > pos){
            len = w - pos - 1;
            p1 = pos + 1;
            p2 = w - 1;
            Left = 0, Right = p.size()-1;
            while(Left < Right){
                int Mid = Left + (Right - Left) / 2;
                if(p[Mid] >= p1) Right = Mid;
                else Left = Mid + 1;
            }
            it1 = Left;
            Left = 0, Right = p.size()-1;
            while(Left < Right){
                int Mid = Left + (Right - Left) / 2;
                if(p[Mid] > p2) Right = Mid;
                else Left = Mid + 1;
            }
            it2 = Left;
            if(p[it1] >= p1 && p[it2] > p2){
                ext = it2 - it1;
//                printf("%I64d\n", ext);
                ans = ans + len - ext;
            }
        }
    }
    printf("%I64d\n", ans);
}