BZOJ 3932 [CQOI2015]任务查询系统 - 差分 + 主席树

Solution

差分就好了， 在$s_i$ 的点+1, $e_i + 1$ 的点 - 1。

查询的时候注意$l == r$ 要返回 $k * b[l]$ ，而不是$sum[node] $因为当前位置的个数可能大于$k$

最后再心疼自己因为神奇建树挂死2333， 被大佬喷了（

Code

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rd read()
 #define ll long long
 using namespace std;

 const int N = 2e5 + 1e3;

 int n, m, num;
 int b[N], tot, nd_num;
 int lson[N * ], rson[N * ], root[N << ], cnt[N * ];
 ll lastans = , sum[N * ];

 struct node {
     int d, pos, x;
 }a[N << ];

 int read() {
     int X =  , p = ; char c= getchar();
     for( ;c > '' || c < ''; c = getchar()) if( c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 int cmp(const node &A, const node &B) {
     return A.x < B.x;
 }

 int fd(int x) {
     return lower_bound(b + , b +  + tot, x) - b;
 }

 void change(int last, int &now, int pos, int d, int l, int r) {
     now = ++nd_num;
     sum[now] = sum[last] + d * b[pos];
     cnt[now] = cnt[last] + d;
     lson[now] = lson[last];
     rson[now] = rson[last];
     if(l == r) return;
     int mid = (l + r) >> ;
     if(pos <= mid)
         change(lson[last], lson[now], pos, d, l, mid);
     else
         change(rson[last], rson[now], pos, d, mid + , r);
 }

 ll query(int now, int k, int l, int r) {
     if(cnt[now] < k) return sum[now];
     if(l == r)
         return b[l] * k;
     int mid = (l + r) >> , tmp;
     if((tmp = cnt[lson[now]]) >= k)
         return query(lson[now], k, l, mid);
     else
         return sum[lson[now]] + query(rson[now], k - tmp, mid + , r);
 }

 int main()
 {
     n = rd; m = rd;
     for(int i = ; i <= n; ++i) {
         int l = rd, r = rd, pos = rd;
         a[++num].d = ;
         a[num].pos = pos;
         a[num].x = l;
         a[++num].d = -;
         a[num].pos = pos;
         a[num].x = r + ;
         b[++tot] = pos;
     }
     sort(b + , b +  + tot);
     tot = unique(b + , b +  + tot) - b - ;
     sort(a + , a +  + num, cmp);
     for(int i = , j = ; i <= m; ++i) {
                 root[i] = root[i - ];
         for(; j <= num && a[j].x == i; ++j) {
             change(root[i], root[i], fd(a[j].pos), a[j].d, , tot);
         }
     }
     for(int i = ; i <= m; ++i) {
         int x = rd, A = rd, B = rd, C = rd, k;
         k =  + (A * lastans + B) % C;
         lastans = query(root[x], k, , tot);
         printf("%lld\n", lastans);
     }
 }