[ZJOI 2013] bzoj3110 K大数查询 【树套树】
发布时间:2021-01-16 18:02:18 所属栏目:大数据 来源:网络整理
导读:副标题#e# Description 有N个位置,M个操作。操作有两种,每次操作如果是1 a b c的形式表示在第a个位置到第b个位置,每个位置加入一个数c 如果是2 a b c形式,表示询问从第a个位置到第b个位置,第C大的数是多少。 Input 第一行N,M 接下来M行,每行形如1 a
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 50000 + 10;
struct opt {
int i,k,c;
bool operator < (const opt& rhs) const {
if(c == rhs.c) return i < rhs.i;
return c > rhs.c;
}
}A[maxn];
vector<opt> T;
int ID[maxn];
bool ins[maxn];
int main()
{
int n,m;
scanf("%d %d",&n,&m);
for(int i = 1; i <= m; i++) {
int k,c;
scanf("%d %d %d %d",&k,&a,&b,&c);
A[i] = (opt){i,c};
if(k == 1) T.push_back(A[i]);
}
sort(T.begin(),T.end());
for(int i = 0; i < T.size(); i++) ID[T[i].i] = i;
for(int i = 1; i <= m; i++)
if(A[i].k == 1) ins[ID[i]] = 1;
else {
int j,c = 0;
for(j = 0; j < T.size(); j++) if(ins[j]) {
int len = min(A[i].b,T[j].b) - max(A[i].a,T[j].a) + 1;
if(len <= 0) continue;
if((c += len) >= A[i].c) break;
}
printf("%dn",T[j].c);
}
return 0;
}
整体二分 http://blog.csdn.net/qq_21841245/article/details/44906735 #include "cstdio"
#define lowbit(x) (x & (-x))
using namespace std;
const int Nmax = 50005;
int N,M;
struct Option{
int sign,y,c;
}op[Nmax];
int tot = -1,ans[Nmax];
int q[Nmax],tmp[2][Nmax];
namespace BIT{
int t[Nmax][2],d[Nmax][2];
void update(bool s,int pos,int c)
{
for (int i = pos; i <= N; i += lowbit(i)) {
if (t[i][s] != tot) { t[i][s] = tot; d[i][s] = 0; }
d[i][s] += c;
}
}
int get_sum(bool s,int pos)
{
int res = 0;
for (int i = pos; i; i -= lowbit(i)) {
if (t[i][s] != tot) { t[i][s] = tot; d[i][s] = 0; }
res += d[i][s];
}
return res;
}
void Add(int x,int y)
{
update(0,1); update(0,y + 1,-1);
update(1,x); update(1,-(y + 1));
}
int Query(int x,int y)
{
int temp = get_sum(0,y) * (y + 1) - get_sum(1,y);
temp -= get_sum(0,x - 1) * x - get_sum(1,x - 1);
return temp;
}
}
void solve(int L,int r)
{
if (L > R) return;
++tot; int mid = (l + r) >> 1;
if (l == r) {
for (int i = L; i <= R; ++i) if (op[q[i]].sign == 2) ans[q[i]] = mid;
return;
}
tmp[0][0] = tmp[1][0] = 0; using namespace BIT;
for (int i = L; i <= R; ++i) {
int temp = q[i];
if (op[temp].sign == 1) {
if (op[temp].c <= mid) tmp[0][++tmp[0][0]] = temp;
else {
tmp[1][++tmp[1][0]] = temp;
Add(op[temp].x,op[temp].y);
}
} else {
int cnt = Query(op[temp].x,op[temp].y);
if (cnt < op[temp].c) {
op[temp].c -= cnt;
tmp[0][++tmp[0][0]] = temp;
} else tmp[1][++tmp[1][0]] = temp;
}
}
int tl = L,t2 = L + tmp[0][0] - 1;
for (int i = 1; i <= tmp[0][0]; ++i) q[tl++] = tmp[0][i];
for (int i = 1; i <= tmp[1][0]; ++i) q[tl++] = tmp[1][i];
solve(L,t2,mid); solve(t2 + 1,r);
}
int main()
{
scanf("%d%d",&N,&M);
for (int i = 1; i <= M; ++i) {
scanf("%d%d%d%d",&op[i].sign,&op[i].x,&op[i].y,&op[i].c);
q[i] = i;
}
solve(1,M,N);
for (int i = 1; i <= M; ++i) {
if (op[i].sign == 2) printf("%dn",ans[i]);
}
return 0;
}
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