数据结构
栈
队列
链表与邻接表
Hash
字符串
KMP算法
Trie
二叉堆
并查集
树状数组
线段树
#include<cstdlib>
#include<cstring>
#include<iostream>
#define ll long long
using namespace std;
const int MAX = 100000 + 5;
inline int read(){
int f = 1, x = 0;char ch;
do { ch = getchar(); if (ch == '-') f = -1; } while (ch < '0'||ch>'9');
do {x = x*10+ch-'0'; ch = getchar(); } while (ch >= '0' && ch <= '9');
return f*x;
}
int n, m, a[MAX];
struct sakura {
int l, r;
int dat;
ll sum;
ll add;
}sak[4*MAX];
inline void build(int p, int l, int r) {
sak[p].l = l, sak[p].r = r;
if (l == r) {
sak[p].dat = a[l], sak[p].sum = a[l];
return;
}
int mid = (l + r) / 2;
build(2*p, l, mid);
build(2*p + 1, mid + 1, r);
sak[p].sum = sak[2*p].sum + sak[2*p + 1].sum;
}
inline void pushdown(int p) {
if (sak[p].add) {
sak[2*p].sum += sak[p].add*(sak[2*p].r - sak[2*p].l+1);
sak[2*p+1].sum += sak[p].add*(sak[2*p+1].r - sak[2*p+1].l+1);
sak[2*p].add += sak[p].add;
sak[2*p+1].add += sak[p].add;
sak[p].add = 0;
}
}
inline void change(int p, int l, int r, int d) {
if (l <= sak[p].l && r >= sak[p].r) {
sak[p].sum += (ll)d*(sak[p].r - sak[p].l+1);
sak[p].add += d;
return;
}
pushdown(p);
int mid = (sak[p].l + sak[p].r) / 2;
if (l <= mid) change(2*p, l, r, d);
if (r > mid) change(2*p + 1, l, r, d);
sak[p].sum = sak[2*p].sum + sak[2*p + 1].sum;
}
inline ll ask(int p, int l, int r) {
if (l <= sak[p].l && r >= sak[p].r) {
return sak[p].sum;
}
pushdown(p);
ll val = 0;
int mid = (sak[p].l + sak[p].r) / 2;
if (l <= mid) val += ask(2*p, l, r);
if (r > mid) val += ask(2*p + 1, l, r);
return val;
}
int main(){
n = read(), m = read();
for (int i = 1;i <= n; ++i) a[i] = read();
build(1, 1, n);
for (int i = 1;i <= m; ++i) {
int opt = read();
switch (opt) {
case 1:{
int x = read(), y = read(), k = read();
change(1, x, y, k);
break;
}
case 2:{
int x = read(), y = read();
printf("%lld\n", ask(1, x, y));
break;
}
}
}
return 0;
}
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debug() puts("FBI WARNING!")
#define ll long long
using namespace std;
const int MAX = 100000 + 5;
inline int read(){
int f = 1, x = 0;char ch;
do { ch = getchar(); if (ch == '-') f = -1; } while (ch < '0'||ch>'9');
do {x = x*10+ch-'0'; ch = getchar(); } while (ch >= '0' && ch <= '9');
return f*x;
}
int n, m, mod, a[MAX];
struct sakura {
int l, r;
ll sum; ll add; ll mul;
sakura() {
l = 0, r = 0, sum = 0, add = 0, mul = 1;
}
}sak[10*MAX];
inline void build(int p, int l, int r) {
sak[p].l = l, sak[p].r = r;
if (l == r) {
sak[p].sum = a[l]%mod;
return;
}
int mid = (l + r) / 2;
build(2*p, l, mid);
build(2*p + 1, mid + 1, r);
sak[p].sum = (sak[2*p].sum + sak[2*p + 1].sum)%mod;
}
inline void pushdown(int p) {
sak[2*p].sum = (sak[2*p].sum*sak[p].mul + sak[p].add*(sak[2*p].r - sak[2*p].l+1))%mod;
sak[2*p+1].sum = (sak[2*p+1].sum*sak[p].mul + sak[p].add*(sak[2*p+1].r - sak[2*p+1].l+1))%mod;
sak[2*p].add = (sak[p].add + sak[2*p].add*sak[p].mul)%mod;
sak[2*p+1].add = (sak[p].add + sak[2*p+1].add*sak[p].mul)%mod;
sak[2*p].mul = (sak[2*p].mul*sak[p].mul)%mod;
sak[2*p+1].mul = (sak[2*p+1].mul*sak[p].mul)%mod;
sak[p].add = 0;
sak[p].mul = 1;
}
inline void change(int p, int l, int r, ll d) {
if (l <= sak[p].l && r >= sak[p].r) {
sak[p].sum = (sak[p].sum + d*(sak[p].r - sak[p].l+1))%mod;
sak[p].add = (sak[p].add + d)%mod;
return;
}
pushdown(p);
int mid = (sak[p].l + sak[p].r) / 2;
if (l <= mid) change(2*p, l, r, d);
if (r > mid) change(2*p + 1, l, r, d);
sak[p].sum = (sak[2*p].sum + sak[2*p + 1].sum)%mod;
}
inline void mull(int p, int l, int r, int d) {
if (l <= sak[p].l && r >= sak[p].r) {
sak[p].mul = (sak[p].mul * d)%mod;
sak[p].add = (sak[p].add * d)%mod;
sak[p].sum = (sak[p].sum * d)%mod;
return;
}
pushdown(p);
int mid = (sak[p].l + sak[p].r) / 2;
if (l <= mid) mull(2*p, l, r, d);
if (r > mid) mull(2*p + 1, l, r, d);
sak[p].sum = (sak[2*p].sum + sak[2*p + 1].sum)%mod;
}
inline ll ask(int p, int l, int r) {
if (l <= sak[p].l && r >= sak[p].r) {
return sak[p].sum%mod;
}
pushdown(p);
ll val = 0;
int mid = (sak[p].l + sak[p].r) / 2;
if (l <= mid) val = (val + ask(2*p, l, r))%mod;
if (r > mid) val = (val + ask(2*p + 1, l, r))%mod;
return val;
}
int main(){
n = read(), m = read(), mod = read();
for (int i = 1;i <= n; ++i) a[i] = read();
build(1, 1, n);
for (int i = 1;i <= m; ++i) {
int opt = read();
switch (opt) {
case 1:{
int x = read(), y = read(), k = read();
mull(1, x, y, k);
break;
}
case 2:{
int x = read(), y = read(), k = read();
change(1, x, y, k);
break;
}
case 3:{
int x = read(), y = read();
printf("%lld\n", ask(1, x, y));
break;
}
}
}
return 0;
}
#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debug() puts("FBI WARNING!")
#define ll long long
using namespace std;
const int MAX = 1000 + 5;
int n, tot, cnts[4*MAX];
double x1, x2, y, y2, raw[2*MAX], ans;
inline int read(){
char ch = getchar();
int res = 0, f = 1;
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar(); }
while(isdigit(ch)) res = (res<<3) + (res<<1) + (ch^48), ch = getchar();
return res*f;
}
inline void write(ll x){if(x > 9) write(x/10); putchar(x%10+48); return;}
struct sakura{ double x, y1, y2; int k; }sak[MAX]; int cnt;
struct Segment{ double l, r; double sum; }tree[4*MAX];
inline void push_up(int p) {
if (cnts[p] > 0) { tree[p].sum = tree[p].r - tree[p].l; }
else { tree[p].sum = tree[p << 1].sum + tree[(p << 1) + 1].sum; }
}
inline void build(int p, int l, int r) {
if(r-l == 1) {
tree[p].l = raw[l], tree[p].r = raw[r], tree[p].sum = 0;
return;
}
tree[p].l = raw[l], tree[p].r = raw[r];
int mid = (l + r) >> 1;
build(2*p, l, mid);
build(2*p + 1, mid, r);
push_up(p);
return ;
}
inline void change(int p, double y1, double y2, int flag) {
if (tree[p].l == y1 && tree[p].r == y2) {
cnts[p] += flag;
push_up(p);
return;
}
if (tree[p << 1].r > y1) {
change(p << 1, y1, min(y2, tree[p << 1].r), flag);
}
if (tree[(p << 1) + 1].l < y2) {
change((p << 1) + 1, max(tree[(p << 1) + 1].l, y1), y2, flag);
}
push_up(p);
}
inline void clean() {
memset(cnts, 0, sizeof (cnts));
cnt = 0;
ans = 0;
}
inline void add() {
sak[cnt].x = x1, sak[cnt].y1 = y, sak[cnt].y2 = y2, sak[cnt].k = 1;
++cnt;
sak[cnt].x = x2, sak[cnt].y1 = y, sak[cnt].y2 = y2, sak[cnt].k = -1;
++cnt;
}
inline bool cmp(const sakura &a, const sakura &b) { return a.x < b.x; }
int main(){
while (n = read()) {
if (n == 0) break;
clean();
for (int i = 1;i <= n; ++i) {
scanf("%lf%lf%lf%lf", &x1, &y, &x2, &y2);
add();
raw[i] = y;
raw[i+n] = y2;
}
sort(sak, sak+2*n, cmp);
sort(raw+1, raw+(2*n+1));
build(1, 1, n << 1);
change(1, sak[0].y1, sak[0].y2, sak[0].k);
for (int i = 1;i < (n << 1); ++i) {
ans += (sak[i].x - sak[i-1].x) * tree[1].sum;
change(1, sak[i].y1, sak[i].y2, sak[i].k);
}
printf("Test case #%d\nTotal explored area: %.2lf\n\n", ++tot, ans);
}
return 0;
}
分块
点分治
BFS
离线分治算法
可持续化数据结构
浙公网安备 33010602011771号