线段树,树状数组专题
2019.7.16
线段树,树状数组专题
HDU 1166 敌兵布阵
单点更新,区间求和,树状数组模板题
#include <cstdio>
#include <cstring>
int bit[50010], n;
void add(int i, int x) {
while(i <= n) {
bit[i] += x;
i += i & -i;
}
}
int sum(int y) {
int s = 0;
while(y > 0) {
s += bit[y];
y -= y & -y;
}
return s;
}
int main() {
int t;
scanf("%d", &t);
for(int k = 1; k <= t; k++) {
memset(bit, 0, sizeof(bit));
printf("Case %d:\n", k);
int a;
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a);
add(i, a);
}
char b[10];
int c, d;
while(1) {
scanf("%s", b);
if(b[0] == 'E') break;
else scanf("%d %d", &c, &d);
if(b[0] == 'Q') printf("%d\n", sum(d) - sum(c-1));
else if(b[0] == 'A') add(c, d);
else if(b[0] == 'S') add(c, -d);
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers
区间修改(加或减某个数),区间求和,树状数组模板题
#include <cstdio>
#define lowbit(i) (i & (-i))
typedef long long LL;
const LL Nmax = 200100;
LL N, Q;
LL delta[Nmax];
LL deltai[Nmax];
LL sum[Nmax];
LL Query(LL pos)
{
LL temp = 0, tempp = 0, poss = pos + 1;
while(pos > 0)
{
temp += delta[pos];
tempp += deltai[pos];
pos -= lowbit(pos);
}
return temp * poss - tempp;
}
void Update(LL pos, LL x)
{
LL poss = x * pos;
while(pos <= N)
{
delta[pos] += x;
deltai[pos] += poss;
pos += lowbit(pos);
}
}
int main()
{
scanf("%lld %lld", &N, &Q);
LL x;
for(LL i = 1; i <= N; ++i)
{
scanf("%lld", &x);
sum[i] = sum[i - 1] + x;
}
while(Q--)
{
char sign[10];
LL l, r, x;
scanf("%s", sign);
if(sign[0] == 'C')
{
scanf("%lld %lld %lld", &l, &r, &x);
Update(l, x);
Update(r+1, -x);
}
else
{
scanf("%lld %lld", &l, &r);
LL suml = sum[l - 1] + Query(l - 1);
LL sumr = sum[r] + Query(r);
printf("%lld\n", sumr - suml);
}
}
return 0;
}
HDU 4027 Can you answer these queries?(线段树)
有两种操作。第一种是对某个区间的数开根,第二种操作是查询某区间的和。
单点更新,区间求和,适当剪枝
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 100000+50 ;
ll sum[4*N], add[4*N] ;
void build(int l, int r, int i)
{
sum[i] = 0;
if(l == r)
{
scanf("%lld", &sum[i]);
return ;
}
int mid = (l+r)>>1;
build(l,mid,i<<1);
build(mid+1,r,i<<1|1);
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
void update_sqrt(int i, int l, int r, int ql, int qr ) //更新区间为qlqr,当前区间为l,r,代表当前区间和的节点为i,更新值为val,
{
if(l > qr || ql > r)
return ;
if (l>=ql&&r<=qr&&sum[i]<=r-l+1)
return ;
if(l==r)
{
sum[i]=sqrt(1.0*sum[i]);
return ;
}
int mid = (l + r) >> 1;
update_sqrt(i << 1, l, mid, ql, qr );
update_sqrt(i << 1 | 1, mid + 1, r, ql, qr);
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
ll query(int i, int l, int r, int ql, int qr) //查询区间为qlqr,当前区间为l,r,代表当前区间和的节点为i
{
if(l > qr || ql > r)
return 0;
if(l >= ql && r <= qr)
return sum[i];
int mid =( l + r) >> 1;
return query(i << 1, l, mid, ql, qr)
+ query(i << 1 | 1, mid + 1, r, ql, qr);
}
int main()
{
int n,m;
int cnt=1;
while(scanf("%d", &n)!=EOF)
{
printf("Case #%d:\n",cnt++);
build(1,n,1);
int m;
cin>>m;
int op;
int a,b,c;
for( int i = 1; i <= m; i++)
{
scanf("%d",&op);
if (op==1)
{
scanf("%d%d",&a,&b);
if (a>b)swap(a,b);
printf("%lld\n",query(1, 1,n,a,b) );
}
else if (op==0)
{
scanf("%d%d",&a,&b);
if (a>b)swap(a,b);
update_sqrt(1,1,n,a,b);
}
}
printf("\n");
}
return 0;
}
HDU 1540
https://blog.csdn.net/qq_37383726/article/details/78321600
HDU 3974
https://blog.csdn.net/aqa2037299560/article/details/82829151