K大数查询
本来是刷整体二分的，被这个sb题折腾了一下午，用cin就RE, 用scanf就过了＝_=
收获就是偶然学到了树状数组区间修改区间查询的写法吧。。。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 using namespace std;
5 #define LL long long
6 const int maxn = 1e5 + 10;
7 const int maxq = 5e4 + 10;
8 struct Qry{
9 int a, b;
10 LL c;
11 int id;
12 Qry(int a = 0, int b = 0, LL c = 0, int id = 0):
13 a(a), b(b), c(c), id(id){}
14 }q[maxq], q1[maxq], q2[maxq];
15
16
17 struct Seg{
18 LL sum[maxn<<2], add[maxn<<2];
19 void init(){
20 memset(add, 0, sizeof add);
21 memset(sum, 0, sizeof sum);
22 }
23 void pushup(int rt){
24 sum[rt] = sum[rt<<1] + sum[rt<<1|1];
25 }
26 void pushdown(int rt, int m){
27 if(add[rt]){
28 sum[rt<<1] += (m - (m>>1)) * add[rt];
29 sum[rt<<1|1] += (m >> 1) * add[rt];
30 add[rt<<1] += add[rt];
31 add[rt<<1|1] += add[rt];
32 add[rt] = 0;
33 }
34 }
35
36 void update(int L, int R, int v, int l, int r, int rt){
37 if(L <= l && r <= R){
38 sum[rt] += v * (r - l + 1);
39 add[rt] += v;
40 return;
41 }
42 pushdown(rt, r - l + 1);
43 int m = (l + r) >> 1;
44 if(L <= m) update(L, R, v, l, m, rt<<1);
45 if(R > m) update(L, R, v, m + 1, r, rt<<1|1);
46 pushup(rt);
47 }
48
49 LL query(int L, int R, int l, int r, int rt){
50 if(L <= l && r <= R){
51 return sum[rt];
52 }
53 pushdown(rt, r - l + 1);
54 int m = (l + r) >> 1;
55 LL res = 0;
56 if(L <= m) res += query(L, R, l, m, rt<<1);
57 if(R > m) res += query(L, R, m + 1, r, rt<<1|1);
58 return res;
59 }
60 }seg;
61 int ans[maxq];
62 int n;
63 void solve(int L, int R, int l, int r){
64 if(L > R) return;
65 if(l == r){
66 for(int i = L; i <= R; i++){
67 if(q[i].id > 0) ans[q[i].id] = l;
68 }
69 return;
70 }
71 int m = (l + r) >> 1;
72 int f = 0, g = 0;
73 for(int i = L; i <= R; i++){
74 if(q[i].id < 0){
75 if(q[i].c <= m){
76 seg.update(q[i].a, q[i].b, 1, 1, n, 1);
77 q1[f++] = q[i];
78 }else q2[g++] = q[i];
79 }else{
80 LL temp = seg.query(q[i].a, q[i].b, 1, n, 1);
81 if(temp >= q[i].c) q1[f++] = q[i];
82 else{
83 q[i].c -= temp;
84 q2[g++] = q[i];
85 }
86 }
87 }
88 for(int i = 0; i < f; i++) if(q1[i].id < 0) seg.update(q1[i].a, q1[i].b, -1, 1, n, 1);
89 memcpy(q + L, q1, f * sizeof(Qry));
90 memcpy(q + L + f, q2, g * sizeof(Qry));
91 solve(L, L + f - 1, l, m);
92 solve(L + f, R, m + 1, r);
93 }
94
95 int main(){
96 int m;
97 while(scanf("%d %d", &n, &m) != EOF){
98 int op, a, b;
99 LL c;
100 seg.init();
101 int cnt = 0;
102 for(int i = 1; i <= m; i++){
103 cin>>op>>a>>b>>c;
104 if(op == 1){
105 q[i] = Qry(a, b, n + 1LL - c, -1);
106 }else{
107 q[i] = Qry(a, b, c, ++cnt);
108 }
109 }
110 solve(1, m, 1, n * 2LL + 1);
111 for(int i = 1; i <= cnt; i++) printf("%d\n", n - ans[i] + 1);
112 }
113 }
线段树维护
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 using namespace std;
5 #define LL long long
6 const int maxn = 1e5 + 10;
7 const int maxq = 1e5 + 10;
8 struct Qry{
9 int a, b;
10 LL c;
11 int id;
12 Qry(int a = 0, int b = 0, LL c = 0, int id = 0):
13 a(a), b(b), c(c), id(id){}
14 }q[maxq], q1[maxq], q2[maxq];
15 //树状数组区间修改区间求和 http://blog.csdn.net/blackjack_/article/details/74997479
16 struct Bit{
17 int n;
18 LL c1[maxn], c2[maxn];
19 void init(int _n){
20 n = _n;
21 memset(c1, 0, sizeof(c1));
22 memset(c2, 0, sizeof(c2));
23 }
24 void add(int x,LL add){
25 for(int i=x;i<=n;i+=i&(-i)){
26 c1[i]+=add;
27 c2[i]+=x*add;
28 }
29 }
30 LL sum(int x){
31 LL ans=0;
32 for(int i=x;i>0;i-=i&(-i)){
33 ans+=(x+1)*c1[i]-c2[i];
34 }
35 return ans;
36 }
37 }bit;
38 int ans[maxq];
39 int n;
40 void solve(int L, int R, int l, int r){
41 if(L > R) return;
42 if(l == r){
43 for(int i = L; i <= R; i++){
44 if(q[i].id > 0) ans[q[i].id] = l;
45 }
46 return;
47 }
48 int m = (l + r) >> 1;
49 int f = 0, g = 0;
50 for(int i = L; i <= R; i++){
51 if(q[i].id < 0){
52 if(q[i].c <= m){
53 bit.add(q[i].a, 1LL);
54 bit.add(q[i].b+1, -1LL);
55 q1[f++] = q[i];
56 }else q2[g++] = q[i];
57 }else{
58 LL temp = bit.sum(q[i].b) - bit.sum(q[i].a - 1);
59 if(temp >= q[i].c) q1[f++] = q[i];
60 else{
61 q[i].c -= temp;
62 q2[g++] = q[i];
63 }
64 }
65 }
66 for(int i = 0; i < f; i++) if(q1[i].id < 0 && q1[i].c <= m) {
67 bit.add(q1[i].a, -1);
68 bit.add(q1[i].b + 1, 1);
69 }
70 memcpy(q + L, q1, f * sizeof(Qry));
71 memcpy(q + L + f, q2, g * sizeof(Qry));
72 solve(L, L + f - 1, l, m);
73 solve(L + f, R, m + 1, r);
74 }
75
76 int main(){
77 int m;
78 while(scanf("%d %d", &n, &m)!= EOF){
79 int op, a, b;
80 LL c;
81 bit.init(n);
82 int cnt = 0;
83 for(int i = 1; i <= m; i++){
84 cin>>op>>a>>b>>c;
85 if(op == 1){
86 q[i] = Qry(a, b, n + 1LL - c, -1);
87 }else{
88 q[i] = Qry(a, b, c, ++cnt);
89 }
90 }
91 solve(1, m, 1, n * 2LL + 1);
92 for(int i = 1; i <= cnt; i++) printf("%d\n", n - ans[i] + 1);
93 }
94 }
树状数组维护
原文地址：https://www.cnblogs.com/yijiull/p/8324723.html
时间： 2024-10-06 05:03:42