BZOJ 3110：[Zjoi2013]K大数查询（整体二分）

http://www.lydsy.com/JudgeOnline/problem.php?id=3110

题意：……

思路：其实和之前POJ那道题差不多，只不过是换成区间更新，而且是第k大不是第k小，第k大是降序的第k个，在二分询问的时候需要注意和第k小的不同细节。

树状数组比线段树快了几倍，所以说树状数组区间更新区间查询是一个值得学的姿势啊。

看了一下别人的，可以把插入时候的值换成n - x + 1，那么就可以变成找第k小了，输出的时候还要换回来。

线段树：

 //9028 kb    7484 ms
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define N 50010
 #define lson rt<<1, l, m
 #define rson rt<<1|1, m + 1, r
 typedef long long LL;
 struct P {
     int type, l, r, id; LL val;
     P () {}
     P (int type, int l, int r, LL val, int id) : type(type), l(l), r(r), val(val), id(id) {}
 } q[N], lq[N], rq[N], qq[N];
 LL tree[N<<], lazy[N<<], ans[N];
 int n;

 void pushup(int rt) { tree[rt] = tree[rt<<|] + tree[rt<<]; }

 void pushdown(int rt, int len) {
     if(lazy[rt]) {
         lazy[rt<<] += lazy[rt]; lazy[rt<<|] += lazy[rt];
         tree[rt<<] += lazy[rt] * (len - (len >> )); tree[rt<<|] += lazy[rt] * (len >> );
         lazy[rt] = ;
     }
 }

 void update(int rt, int l, int r, int L, int R, int val) {
     if(L <= l && r <= R) {
         tree[rt] += (r - l + ) * val; // 居然漏了+1
         lazy[rt] += val; return ;
     }
     pushdown(rt, r - l + );
     int m = (l + r) >> ;
     if(L <= m) update(lson, L, R, val);
     if(m < R) update(rson, L, R, val);
     pushup(rt);
 }

 LL query(int rt, int l, int r, int L, int R) {
     if(L <= l && r <= R) return tree[rt];
     pushdown(rt, r - l + );
     int m = (l + r) >> ;
     LL ans = ;
     if(L <= m) ans += query(lson, L, R);
     if(m < R) ans += query(rson, L, R);
     return ans;
 }

 void Solve(int l, int r, int lask, int rask) {
     if(rask < lask || r < l) return ;
     if(l == r) { for(int i = lask; i <= rask; i++) if(q[i].type == ) ans[q[i].id] = l; return ; }
     int mid = (l + r) >> , lcnt = , rcnt = ;
     for(int i = lask; i <= rask; i++) {
         if(q[i].type == ) {
             if(mid >= q[i].val) { // 第k大 = 降序第k个,只更新比当前二分的答案大的
                 lq[++lcnt] = q[i];
             } else {
                 rq[++rcnt] = q[i];
                 update(, , n, q[i].l, q[i].r, );
             }
         } else {
             LL num = query(, , n, q[i].l, q[i].r);
             if(num >= q[i].val) rq[++rcnt] = q[i];
             else {
                 q[i].val -= num;
                 lq[++lcnt] = q[i];
             }
         }
     }
     for(int i = lask; i <= rask; i++) if(q[i].type ==  && mid < q[i].val) update(, , n, q[i].l, q[i].r, -);
     for(int i = ; i <= lcnt; i++) q[i+lask-] = lq[i];
     for(int i = ; i <= rcnt; i++) q[i+lask+lcnt-] = rq[i];
     Solve(l, mid, lask, lask + lcnt - );
     Solve(mid + , r, lask + lcnt, rask);
 }

 int main() {
     int m, a, b, c, d, ins = , ask = ;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d%d", &a, &b, &c, &d);
         if(a == ) q[i] = P(a, b, c, d, ++ins);
         else q[i] = P(a, b, c, d, ++ask);
     }
     Solve(, n, , m);
     for(int i = ; i <= ask; i++)
         printf("%lld\n", ans[i]);
     return ;
 }

树状数组：

 //1828ms    6.7MB
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define N 50010
 #define lson rt<<1, l, m
 #define rson rt<<1|1, m + 1, r
 typedef long long LL;
 struct P {
     int type, l, r, id; LL val;
     P () {}
     P (int type, int l, int r, LL val, int id) : type(type), l(l), r(r), val(val), id(id) {}
 } q[N], lq[N], rq[N], qq[N];
 LL bit[][N], ans[N];
 int n;

 int lowbit(int x) { return x & (-x); }
 void Add(int i, int x, int w) { while(x <= n) bit[i][x] += w, x += lowbit(x); }
 LL Sum(int i, int x) { LL ans = ; while(x) ans += bit[i][x], x -= lowbit(x); return ans; }
 void update(int l, int r, int w) {
     Add(, l, w); Add(, r + , -w); Add(, l, w * l); Add(, r + , -w * (r + ));
 }
 LL query(int l, int r) {
     LL lsum = Sum(, l - ) * l - Sum(, l - );
     LL rsum = Sum(, r) * (r + ) - Sum(, r);
     return rsum - lsum;
 }

 void Solve(int l, int r, int lask, int rask) {
     if(rask < lask || r < l) return ;
     if(l == r) { for(int i = lask; i <= rask; i++) if(q[i].type == ) ans[q[i].id] = l; return ; }
     int mid = (l + r) >> , lcnt = , rcnt = ;
     for(int i = lask; i <= rask; i++) {
         if(q[i].type == ) {
             if(mid >= q[i].val) { // 第k大 = 降序第k个,只更新比当前二分的答案大的
                 lq[++lcnt] = q[i];
             } else {
                 rq[++rcnt] = q[i];
                 update(q[i].l, q[i].r, );
             }
         } else {
             LL num = query(q[i].l, q[i].r);
             if(num >= q[i].val) rq[++rcnt] = q[i];
             else {
                 q[i].val -= num;
                 lq[++lcnt] = q[i];
             }
         }
     }
     for(int i = lask; i <= rask; i++) if(q[i].type ==  && mid < q[i].val) update(q[i].l, q[i].r, -);
     for(int i = ; i <= lcnt; i++) q[i+lask-] = lq[i];
     for(int i = ; i <= rcnt; i++) q[i+lask+lcnt-] = rq[i];
     Solve(l, mid, lask, lask + lcnt - );
     Solve(mid + , r, lask + lcnt, rask);
 }

 int main() {
     int m, a, b, c, d, ins = , ask = ;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d%d", &a, &b, &c, &d);
         if(a == ) q[i] = P(a, b, c, d, ++ins);
         else q[i] = P(a, b, c, d, ++ask);
     }
     Solve(, n, , m);
     for(int i = ; i <= ask; i++)
         printf("%lld\n", ans[i]);
     return ;
 }

第k小：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define N 50010
 #define lson rt<<1, l, m
 #define rson rt<<1|1, m + 1, r
 typedef long long LL;
 struct P {
     int type, l, r, id; LL val;
     P () {}
     P (int type, int l, int r, LL val, int id) : type(type), l(l), r(r), val(val), id(id) {}
 } q[N], lq[N], rq[N], qq[N];
 LL bit[][N], ans[N];
 int n;

 int lowbit(int x) { return x & (-x); }
 void Add(int i, int x, int w) { while(x <= n) bit[i][x] += w, x += lowbit(x); }
 LL Sum(int i, int x) { LL ans = ; while(x) ans += bit[i][x], x -= lowbit(x); return ans; }
 void update(int l, int r, int w) {
     Add(, l, w); Add(, r + , -w); Add(, l, w * l); Add(, r + , -w * (r + ));
 }
 LL query(int l, int r) {
     LL lsum = Sum(, l - ) * l - Sum(, l - );
     LL rsum = Sum(, r) * (r + ) - Sum(, r);
     return rsum - lsum;
 }

 void Solve(int l, int r, int lask, int rask) {
     if(rask < lask || r < l) return ;
     if(l == r) { for(int i = lask; i <= rask; i++) if(q[i].type == ) ans[q[i].id] = l; return ; }
     int mid = (l + r) >> , lcnt = , rcnt = ;
     for(int i = lask; i <= rask; i++) {
         if(q[i].type == ) {
             if(mid >= q[i].val) { // 第k大 = 降序第k个,只更新比当前二分的答案大的
                 lq[++lcnt] = q[i];
                 update(q[i].l, q[i].r, );
             } else {
                 rq[++rcnt] = q[i];
             }
         } else {
             LL num = query(q[i].l, q[i].r);
             if(num >= q[i].val) lq[++lcnt] = q[i];
             else {
                 q[i].val -= num;
                 rq[++rcnt] = q[i];
             }
         }
     }
     for(int i = lask; i <= rask; i++) if(q[i].type ==  && mid >= q[i].val) update(q[i].l, q[i].r, -);
     for(int i = ; i <= lcnt; i++) q[i+lask-] = lq[i];
     for(int i = ; i <= rcnt; i++) q[i+lask+lcnt-] = rq[i];
     Solve(l, mid, lask, lask + lcnt - );
     Solve(mid + , r, lask + lcnt, rask);
 }

 int main() {
     int m, a, b, c, d, ins = , ask = ;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d%d", &a, &b, &c, &d);
         if(a == ) q[i] = P(a, b, c, n - d + , ++ins);
         else q[i] = P(a, b, c, d, ++ask);
     }
     Solve(,  * n + , , m);
     for(int i = ; i <= ask; i++)
         printf("%lld\n", n - ans[i] + );
     return ;
 }