第一题：二维树状数组，bit[x][y]表示从(1,1)到（x,y）的和，二维的坐标从一维的推过来，正确性可以用一个递增和一个递减的序列单调性证明，至于构图就当黑箱吧
#include <cstdio>
int n, m, q;
struct Case1 {
int bit[100010];
void modify( int p, int d ) {
for( int i = p; i <= m; i += i & -i )
bit[i] += d;
}
int query( int r ) {
int rt = 0;
for( int i = r; i; i -= i & -i )
rt += bit[i];
return rt;
}
int query( int l, int r ) {
return query(r) - query(l-1);
}
void solve() {
while (q--) {
char ss[100];
scanf( "%s", ss );
if( ss[0] == ‘m‘ ) {
int x, y, d;
scanf( "%d%d%d", &x, &y, &d );
modify( y, d );
} else {
int x1, y1, x2, y2;
scanf( "%d%d%d%d", &x1, &y1, &x2, &y2 );
printf( "%d\n", query(y1,y2) );
}
}
}
}case1;//一维
struct Case2 {
int bit[1100][1100];
void modify( int x, int y, int d ) {
for( int i = x; i <= n; i += i & -i )
for( int j = y; j <= m; j += j & -j )
bit[i][j] += d;
}
int query( int x, int y ) {
int rt = 0;
for( int i = x; i; i -= i & -i )
for( int j = y; j; j -= j & -j )
rt += bit[i][j];
return rt;
}
int query( int x1, int y1, int x2, int y2 ) {
return query(x2,y2) - query(x1-1,y2) - query(x2,y1-1) + query(x1-1,y1-1);
}
void solve() {
while(q--) {
char ss[100];
scanf( "%s", ss );
if( ss[0] == ‘m‘ ) {
int x, y, d;
scanf( "%d%d%d", &x, &y, &d );
modify( x, y, d );
} else {
int x1, y1, x2, y2;
scanf( "%d%d%d%d", &x1, &y1, &x2, &y2 );
printf( "%d\n", query(x1,y1,x2,y2) );
}
}
}
}case2;//二维
int main() {
freopen ( "matsum.in", "r", stdin ) ;
freopen ( "matsum.out", "w", stdout ) ;
scanf( "%d%d%d", &n, &m, &q );
if( n == 1 )
case1.solve();
else
case2.solve();
}
第二题：裸的线段树模板，注意把值传上去和放下去的地方，写挂了很多次。。。
#include<bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
#define MAX_N 100005
long long sum ;
int mn, mx, a[MAX_N];
int n, m;
struct SegTree{
struct node{
long long sum;
int vmin,vmax,lazy;
};
node Tree[MAX_N << 2];
#define ls l, m, v << 1
#define rs m+1, r, v << 1 | 1
void push_up(int v){
Tree[v].sum = Tree[v << 1].sum + Tree[v << 1 | 1].sum;
Tree[v].vmin = min(Tree[v << 1].vmin, Tree[v << 1 | 1].vmin);
Tree[v].vmax = max(Tree[v << 1].vmax, Tree[v << 1 | 1].vmax);
}
void push_down(int l, int r, int v){
int m = (l + r) >> 1;
Tree[v << 1].sum += 1LL * Tree[v].lazy * (m - l +1);
Tree[v << 1].vmin += Tree[v].lazy;
Tree[v << 1].vmax += Tree[v].lazy;
Tree[v << 1].lazy += Tree[v].lazy;
Tree[v << 1 | 1].sum += 1LL * Tree[v].lazy * (r - m);
Tree[v << 1 | 1].vmin += Tree[v].lazy;
Tree[v << 1 | 1].vmax += Tree[v].lazy;
Tree[v << 1 | 1].lazy += Tree[v].lazy;
Tree[v].lazy = 0;
}
void build(int l = 1, int r = n, int v = 1){
Tree[v].sum = Tree[v].vmin = Tree[v].vmax = Tree[v].lazy = 0;
if(l ==r){
Tree[v].sum = 1LL * a[l];
Tree[v].vmin = Tree[v].vmax = a[l];
}
else {
int m = (l +r) >> 1;
build(ls);
build(rs);
push_up(v);
}
}
void modify(int x,int L,int R,int l = 1, int r = n, int v = 1){
if(l >= L && r <= R){// += x WA了8次，每次都用的lazy...
Tree[v].lazy += x;
Tree[v].sum += 1LL * x * (r - l + 1);
Tree[v].vmin += x;
Tree[v].vmax += x;
}
else {
if(Tree[v].lazy)push_down(l,r,v);
int m = (l + r) / 2;
if(L <= m)modify(x,L,R,ls);
if(R > m)modify(x,L,R,rs);
push_up(v);
}
}
node query(int L,int R,int l = 1, int r = n,int v = 1){
if(l >= L && r <= R)
return Tree[v];
else {
if(Tree[v].lazy) push_down(l,r,v);
int m = (l + r) / 2;
node ans;
ans.vmin = oo, ans.vmax = -oo, ans.sum = 0;//init
if(L <= m){
node nw = query(L,R,ls);
ans.vmin = min(ans.vmin, nw.vmin);
ans.vmax = max(ans.vmax, nw.vmax);
ans.sum += nw.sum;
}
if(R > m){
node nw = query(L,R,rs);
ans.vmin = min(ans.vmin, nw.vmin);
ans.vmax = max(ans.vmax, nw.vmax);
ans.sum += nw.sum;
}
return ans;
}
}
};
SegTree Tr;
int main(){
freopen("segsum.in","r",stdin);
freopen("segsum.out","w",stdout);
cin>>n>>m;
for(int i = 1; i <= n; i++)
scanf("%d", a + i);
Tr.build();
for(int i = 1; i <= m; i++){
string opt;
cin>>opt;
if(opt[0] == ‘q‘){
int l, r;
scanf("%d%d",&l,&r);
SegTree::node nw;
nw = Tr.query(l,r);
cout<<nw.vmin<<" "<<nw.vmax<<" "<<nw.sum<<endl;
}
else{
int l, r, x;
scanf("%d%d%d",&l,&r,&x);
Tr. modify(x,l,r);
}
}
}
线段树找了好几个板子，还是郭大侠的最好用，写起来也简单，但还是WA了很多次
注意：modify，query，build时要push_down， query时要push_up
modify是增加的val，不是lazy；
要记住更新m的值，ans给初值
不过指针跑的真的快
#include <bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
int a[maxn],n,m,s;
void read(int &x){
int f=1;x=0;char s=getchar();
while(s<‘0‘||s>‘9‘){if(s==‘-‘)f=-1;s=getchar();}
while(s>=‘0‘&&s<=‘9‘){x=x*10+s-‘0‘;s=getchar();}
x*=f;
}
struct Info{
ll sum;
int vmin,vmax;
Info(){}
Info(ll sum,int vimn,int vmax):sum(sum),vmin(vimn),vmax(vmax){}
};
Info Merge(const Info &ls,const Info &rs){
return Info((ls.sum + rs.sum), min(ls.vmin, rs.vmin), max(ls.vmax, rs.vmax) );
}
struct Node{
Info info;
int delta;
Node *ls, *rs;
void pushdown(int l,int r){
if(!delta)return ;//!
int m = (l + r) >> 1;
ls->info.sum += 1LL * (m - l +1) * delta;
ls->info.vmin += delta;
ls->info.vmax += delta;
ls->delta += delta;
rs->info.sum += 1LL *(r - m) * delta;
rs->info.vmin += delta;
rs->info.vmax += delta;
rs->delta += delta;
delta = 0;
}
void update(){//!
info = Merge(ls->info, rs->info);
}
}pool[maxn * 3],*tail = pool, *root;
Node * build(int l = 1,int r = n){
Node *nd = ++tail;
if(l == r){
nd->info = Info(a[l],a[l],a[l]); nd->delta = 0;
}
else {
int m = (l + r) >> 1;
nd->ls = build(l, m);
nd->rs = build(m + 1, r);
nd->delta = 0;
nd->update();
}
return nd;
}
#define Ls nd->ls, l, m
#define Rs nd->rs, m+1, r
void modify(int delta, int L, int R, Node *nd, int l = 1, int r = n){
if(l >= L && r <= R){
nd->info.sum += 1LL * delta * (r - l + 1);
nd->info.vmin += delta;
nd->info.vmax += delta;
nd->delta += delta;
}
else {
nd->pushdown(l, r);
int m = (l + r) >> 1;
if(L <= m)modify(delta, L, R, Ls);
if(R > m)modify(delta, L, R, Rs);
nd->update();
}
}
Info query(int L, int R, Node *nd, int l = 1, int r = n ){
if(l >= L && r <= R)
return nd->info;
else {
nd->pushdown(l, r);
int m = (l + r) >> 1;
if(R <= m)return query(L, R, Ls);
else if(L > m)return query(L, R, Rs);
else return Merge( query( L, R, Ls ),query( L, R, Rs ) );
}
}
int main(){
freopen("segsum.in","r",stdin);
freopen("segsum.out","w",stdout);
read(n), read(m);
for(int i = 1; i <= n; i++)
read(a[i]);
root = build();
for(int i = 1; i <= m; i++){
string opt;
cin>>opt;
if(opt[0] == ‘q‘){
int l, r;
read(l),read(r);
Info nw = query(l,r,root);
cout<<nw.vmin<<" "<<nw.vmax<<" "<<nw.sum<<endl;
}
else{
int l, r, x;
read(l), read(r), read(x);
modify(x,l,r,root);
}
}
}
第三题：利用 gcd 可以取并集的特点，二维st表，将一个矩形分成四部分，求并起来的gcd，
求log是要开到maxn,不是n啊，这个错看了三天，最后还是余力大佬帮忙看出来的
#include<iostream>
#include<cstdio>
using namespace std;
#define RG register//卡常
#define maxn 508
#define P 10
int lo[maxn],d[maxn][maxn][P + 1][P + 1];
int n, m, q;
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a%b) ;
}
void read(int &x){
int f=1;x=0;char s=getchar();
while(s<‘0‘||s>‘9‘){if(s==‘-‘)f=-1;s=getchar();
}
while(s<=‘9‘&&s>=‘0‘){ x=x*10+s-‘0‘;s=getchar();
}
x*=f;
}
void init() {
for( RG int pi = 0; pi <= P; pi++ )
for( RG int pj = 0; pj <= P; pj++ ) {
if( pi == 0 && pj == 0 ) continue;
for( RG int i = 1; i + (1<<pi) - 1 <= n; i++ )
for( RG int j = 1; j + (1<<pj) - 1 <= m; j++ ) {
if( pi == 0 ) {
d[i][j][pi][pj] = gcd( d[i][j][pi][pj-1], d[i][j + (1<<(pj-1))][pi][pj-1] );
} else {
d[i][j][pi][pj] = gcd( d[i][j][pi-1][pj], d[i + (1<<(pi-1))][j][pi-1][pj] );
}
}
}
}
int query(int x1,int yy,int x2,int y2){
int pi = lo[x2 - x1 + 1], pj = lo[y2 - yy + 1];
int ans1 = gcd(d[x1][yy][pi][pj], d[x2 - (1 << pi) + 1][yy][pi][pj]);
int ans2 = gcd(d[x1][y2 - (1 << pj) + 1][pi][pj], d[x2 - (1 << pi) + 1][y2 - (1 << pj) + 1][pi][pj]);
return gcd(ans1, ans2);
}
int main(){
freopen("matgcd.in","r",stdin);
freopen("matgcd.out","w",stdout);
read(n),read(m),read(q);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
read(d[i][j][0][0]);
lo[0] = -1;
for(int i = 1; i < maxn; i++)
lo[i] = lo[i / 2] + 1;
init();
for(int i = 1; i <= q; i++){
string opt;
cin>>opt;
int x1,yy,x2,y2;
read(x1),read(yy),read(x2),read(y2);
printf("%d\n",query(x1,yy,x2,y2));
}
}
原文地址：https://www.cnblogs.com/EdSheeran/p/8423591.html
时间： 2024-10-09 19:21:06