Educational Codeforces Round 36 (Rated for Div. 2) E. Physical Education Lessons

提供两种思路
一种线段树区间更新
另一种用map维护连续的区间，也是题解的思路
第二种很难写(我太渣，看了别人的代码，发现自己写的太烦了)

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 6e5+5;
#define MS(x,y) memset(x,y,sizeof(x))
#define MP(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1

int main() {
    int n, q;
    while(~scanf("%d %d", &n, &q)) {
        map<int, int> mp;
        mp[-1] = -1;
        mp[1] = n;
        mp[n+1] = INF;

        int ans = n;
        while(q --) {
            int l, r, k; 


            scanf("%d %d %d", &l, &r, &k);  
            int tt = r;
            if(k == 2 && r != n) tt = r+1;

            auto it = mp.upper_bound(l);
            auto it2 = mp.upper_bound(tt);
            it --; it2 --;

            int head1 = it -> first;  int len1 = it -> second;
            int head2 = it2 -> first;  int len2 = it2 -> second;

            while(1){
                int flag = 0;
                if(it == it2) flag = 1;
                ans -= it->second;
                auto tmp = it;  
            //  if(it->first == n+1) while(1);
                it ++;
        //      printf("erase: %d\n", tmp->first);
                mp.erase(tmp);
                if(flag) break;
            }
        //for(auto i = mp.begin(); i != mp.end(); ++i) printf("%d:%d ", i->first, i->second); printf("\n");
        //  printf("%d %d %d %d %d %d\n", head1, len1, head2, len2, l, r);

            if(k == 1) {
                if(head1 + len1 - 1 >= l && l!=head1) {
                    mp[head1] = l - head1;
                    ans += l - head1;
                } else if(l != head1){
                    mp[head1] = len1; 
                    ans += len1;
                }

                if(head2 + len2 - 1 > r) {
                    mp[r+1] = head2 + len2 - 1 - r;
                    ans += head2 + len2 - 1 - r;
                }           
            } else {
                int L = l; int R = max(r, head2 + len2 -1);
                if(head1 + len1 < l) { 
                    mp[head1] = len1;
                    ans += len1;
                }else L = head1;

                mp[L] = R-L+1;
                ans += R-L+1;
            }



    //      for(auto i = mp.begin(); i != mp.end(); ++i) printf("%d:%d ", i->first, i->second); printf("\n");

            printf("%d\n", ans);
        }
    }
    return 0;
}

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 6e5+5;
#define MS(x,y) memset(x,y,sizeof(x))
#define MP(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1


int Q[N][3];
int has[N * 2]; int cnt;

int sum[N << 2];
int lazy[N << 2];

void build(int l, int r, int rt) {
    lazy[rt] = 0;
    sum[rt] = has[r] - has[l-1];
    if(l == r) {
        return;
    }
    int m = (l + r) >> 1;
    build(lson); build(rson);
}
void update(int ty, int L, int R, int l, int r, int rt) {
    //printf("%d %d\n", l, r);
    if(L <= has[l-1]+1 && has[r] <= R) {
        lazy[rt] = ty == 1? -1 : 1;
        sum[rt] = ty == 1? 0 : has[r] - has[l-1];
        return;
    }


    int m = (l + r) >> 1;

    if(lazy[rt] == 1) {
        lazy[rt<<1] = 1; sum[rt<<1] = has[m] - has[l-1];
        lazy[rt<<1|1] = 1; sum[rt<<1|1] = has[r] - has[m]; 
        lazy[rt] = 0;
    } else if(lazy[rt] == -1){
        lazy[rt<<1] = -1; sum[rt<<1] = 0;
        lazy[rt<<1|1] = -1; sum[rt<<1|1] = 0;
        lazy[rt] = 0;
    }

    if(L <= has[m-1]+1) update(ty, L, R, lson);
    if(R > has[m])  update(ty, L, R, rson);
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void debug(int l, int r, int rt) {
    printf("%d %d %d\n", l, r, sum[rt]);
    if(l == r) return;
    int m = (l + r) >> 1;
    debug(lson);
    debug(rson);
}
int main() {
    int n;
    while(~scanf("%d", &n)) {
        cnt = 0;

        int q; scanf("%d", &q);
        for(int i = 0; i < q; ++i) {
            scanf("%d %d %d", &Q[i][0], &Q[i][1], &Q[i][2]);
            has[cnt ++] = Q[i][0] - 1; 
            has[cnt ++] = Q[i][1];
        }
        has[cnt ++] = 0;
        has[cnt ++] = n;

        sort(has, has+cnt);
        cnt = unique(has, has + cnt) - has;
//      for(int i = 0; i < cnt; ++i) printf("%d ", has[i]); printf("\n");


        build(1, cnt-1, 1);
    //` debug(1, cnt-1, 1);
        for(int i = 0; i < q; ++i) {
            update(Q[i][2], Q[i][0], Q[i][1], 1, cnt-1, 1);
            printf("%d\n", sum[1]); 
    //      debug(1, cnt-1, 1);
        }

    }
    return 0;
}