树状数组详解与模版

单点更新

void update(int x,int y,int n){
    for(int i=x;i<=n;i+=lowbit(i))    //x为更新的位置,y为加的数,n为数组最大值
        c[i] += y;
}

区间查询（1 - x）

int getsum(int x){
    int ans = 0;
    for(int i=x;i;i-=lowbit(i))
        ans += c[i];
    return ans;
}

高级操作

求逆序对

对于数组a，我们将其离散化处理为b[].区间查询与单点修改代码如下

int lowbit(int x) {
    return x & (-x);
}

void update(int p) {
    while (p <= n) {
        c[p] ++;
        p += lowbit(p);
    }
}

int getsum(int p) {
    int res = 0;
    while(p) {
        res += c[p];
        p -= lowbit(p);
    }
    return res;
}

a的逆序对个数为：

for(int i = 1; i <= n; i++) {
            update(b[i]);
            res += (i - getsum(b[i]));
        }

poj 2299

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<math.h>
#include<vector>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>


using namespace std;
typedef long long LL;
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
const int MAXN = 5e5 + 10;
const double eps = 1e-8;

int c[MAXN];
int n;
int a[MAXN],b[MAXN];

int lowbit(int x) {
    return x & (-x);
}

void update(int p) {
    while (p <= n) {
        c[p] ++;
        p += lowbit(p);
    }
}

int getsum(int p) {
    int res = 0;
    while(p) {
        res += c[p];
        p -= lowbit(p);
    }
    return res;
}
int main()
{

    while (~scanf("%d", &n) && n) {
        memset(c,0,sizeof c);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(a + 1, a + 1 + n);
        int cnt = unique(a + 1, a + 1 + n) - a - 1;
        for (int i = 1; i <= n; i++)
            b[i] = lower_bound(a + 1, a + 1 + cnt, b[i]) - a;

        LL res = 0;
        for(int i = 1; i <= n; i++) {
            update(b[i]);
            res += (i - getsum(b[i]));
        }
        printf("%lld
",res);
    }
}

res就是逆序对个数，ask，需注意b[i]应该大于0

求区间最大值

void Update(int i,int v)
{
    while(i<=maxY)
    {
        t[i] = max(t[i],v);
        i += lowbit(i);
    }
}

int query(int i)
{
    int ans = 0;
    while(i)
    {
        ans = max(ans,t[i]);
        i -= lowbit(i);
    }
    return ans;
}

int lowbit(int x) {
    return x & -x;
}
void add(int p,int x) {    //这个函数用来在树状数组中直接修改
    while(p <= n) {
        c[p] += x;
        p += lowbit(p);
    }
}

void range_add(int l,int r,int x) {  //给区间[l, r]加上x
    add(l,x);
    add(r + 1, -x);
}
int ask(int p) {  //单点查询前缀和
    int res = 0;
    while(p) {
        res += c[p];
        p -= lowbit(p);
    }
    return res;
}

void add(ll p, ll x){
    for(int i = p; i <= n; i += i & -i)
        sum1[i] += x, sum2[i] += x * p;
}
void range_add(ll l, ll r, ll x){
    add(l, x), add(r + 1, -x);
}
ll ask(ll p){
    ll res = 0;
    for(int i = p; i; i -= i & -i)
        res += (p + 1) * sum1[i] - sum2[i];
    return res;
}
ll range_ask(ll l, ll r){
    return ask(r) - ask(l - 1);
}

用这个做区间修改区间求和的题，无论是时间上还是空间上都比带lazy标记的线段树要优。

 

void add(int x, int y, int z){ //将点(x, y)加上z
    int memo_y = y;
    while(x <= n){
        y = memo_y;
        while(y <= n)
            tree[x][y] += z, y += y & -y;
        x += x & -x;
    }
}
int ask(int x, int y){//求左上角为(1,1)右下角为(x,y) 的矩阵和
    int res = 0, memo_y = y;
    while(x){
        y = memo_y;
        while(y)
            res += tree[x][y], y -= y & -y;
        x -= x & -x;
    }
    return res;
}

void add(int x, int y, int z){
    int memo_y = y;
    while(x <= n){
        y = memo_y;
        while(y <= n)
            tree[x][y] += z, y += y & -y;
        x += x & -x;
    }
}
void range_add(int xa, int ya, int xb, int yb, int z){
    add(xa, ya, z);
    add(xa, yb + 1, -z);
    add(xb + 1, ya, -z);
    add(xb + 1, yb + 1, z);
}
int ask(int x, int y){
    int res = 0, memo_y = y;
    while(x){
        y = memo_y;
        while(y)
            res += tree[x][y], y -= y & -y;
        x -= x & -x;
    }
    return res;
}

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
ll read(){
    char c; bool op = 0;
    while((c = getchar()) < '0' || c > '9')
        if(c == '-') op = 1;
    ll res = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        res = res * 10 + c - '0';
    return op ? -res : res;
}
const int N = 205;
ll n, m, Q;
ll t1[N][N], t2[N][N], t3[N][N], t4[N][N];
void add(ll x, ll y, ll z){
    for(int X = x; X <= n; X += X & -X)
        for(int Y = y; Y <= m; Y += Y & -Y){
            t1[X][Y] += z;
            t2[X][Y] += z * x;
            t3[X][Y] += z * y;
            t4[X][Y] += z * x * y;
        }
}
void range_add(ll xa, ll ya, ll xb, ll yb, ll z){ //(xa, ya) 到 (xb, yb) 的矩形
    add(xa, ya, z);
    add(xa, yb + 1, -z);
    add(xb + 1, ya, -z);
    add(xb + 1, yb + 1, z);
}
ll ask(ll x, ll y){
    ll res = 0;
    for(int i = x; i; i -= i & -i)
        for(int j = y; j; j -= j & -j)
            res += (x + 1) * (y + 1) * t1[i][j]
                - (y + 1) * t2[i][j]
                - (x + 1) * t3[i][j]
                + t4[i][j];
    return res;
}
ll range_ask(ll xa, ll ya, ll xb, ll yb){
    return ask(xb, yb) - ask(xb, ya - 1) - ask(xa - 1, yb) + ask(xa - 1, ya - 1);
}
int main(){
    n = read(), m = read(), Q = read();
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            ll z = read();
            range_add(i, j, i, j, z);
        }
    }
    while(Q--){
        ll ya = read(), xa = read(), yb = read(), xb = read(), z = read(), a = read();
        if(range_ask(xa, ya, xb, yb) < z * (xb - xa + 1) * (yb - ya + 1))
            range_add(xa, ya, xb, yb, a);
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++)
            printf("%lld ", range_ask(i, j, i, j));
        putchar('
');
    }
    return 0;
}