Fast Matrix Operations UVA - 11992 线段树
题意翻译
有一个r行c列的全0矩阵，有以下三种操作。
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1 X1 Y1 X2 Y2 v 子矩阵（X1,Y1,X2,Y2）的元素加v
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2 X1 Y1 X2 Y2 v 子矩阵（X1,Y1,X2,Y2）的元素变为v
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3 X1 Y1 X2 Y2 查询子矩阵（X1,Y1,X2,Y2）的和，最大值，最小值
子矩阵（X1,Y1,X2,Y2）满足X1<=X<=X2 Y1<=Y<=Y2的所有元素（X1,Y2）。
输入保证和不超过10^9
感谢@Himself65 提供的翻译
题目描述
输入输出格式
输入格式：
输入输出样例
输入样例#1：
复制
4 4 8 1 1 2 4 4 5 3 2 1 4 4 1 1 1 3 4 2 3 1 2 4 4 3 1 1 3 4 2 2 1 4 4 2 3 1 2 4 4 1 1 1 4 3 3
输出样例#1： 复制
45 0 5 78 5 7 69 2 7 39 2 7
注意setv和addv的优先级，当有setv时，addv就不应该再有影响；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
struct node {
int sum[maxn], minn[maxn], maxx[maxn], ls[maxn], rs[maxn], setv[maxn], addv[maxn];
int tt, rt;
void init() {
tt = rt = 1; mclr(setv, -1); ms(addv);
ms(sum); ms(minn); ms(maxx); ms(ls); ms(rs);
}
void pushup(int o) {
sum[o] = sum[ls[o]] + sum[rs[o]];
minn[o] = min(minn[ls[o]], minn[rs[o]]);
maxx[o] = max(maxx[ls[o]], maxx[rs[o]]);
}
void pushdown(int o,int l,int r) {
if (setv[o]!=-1) {
int mid = (l + r) >> 1;
if (!ls[o])ls[o] = ++tt; if (!rs[o])rs[o] = ++tt;
sum[ls[o]] = setv[o] * (mid - l + 1);
sum[rs[o]] = setv[o] * (r - mid);
minn[ls[o]] = minn[rs[o]] = maxx[ls[o]] = maxx[rs[o]] = setv[o];
setv[ls[o]] = setv[o]; setv[rs[o]] = setv[o]; setv[o] = -1; addv[ls[o]] = addv[rs[o]] = 0;
}
if (addv[o]) {
int mid = (l + r) >> 1;
if (!ls[o])ls[o] = ++tt; if (!rs[o])rs[o] = ++tt;
sum[ls[o]] += addv[o] * (mid - l + 1); sum[rs[o]] += addv[o] * (r - mid);
maxx[ls[o]] += addv[o]; maxx[rs[o]] += addv[o];
minn[ls[o]] += addv[o]; minn[rs[o]] += addv[o];
addv[ls[o]] += addv[o]; addv[rs[o]] += addv[o];
addv[o] = 0;
}
}
void upd(int &o, int L, int R, int l, int r, int opt,int val) {
if (!o) {
o = ++tt;
}
if (L <= l && r <= R) {
if (opt == 1) {
sum[o] += (r - l + 1)*val;
maxx[o] += val; minn[o] += val;
addv[o] += val;
}
else {
sum[o] = (r - l + 1)*val;
maxx[o] = val; minn[o] = val;
setv[o] = val; addv[o] = 0;
}
return;
}
pushdown(o, l, r);
int mid = (l + r) >> 1;
if (L <= mid)upd(ls[o], L, R, l, mid, opt, val);
if (mid < R)upd(rs[o], L, R, mid + 1, r, opt, val);
pushup(o);
}
int Sum(int L, int R, int l, int r, int o) {
if (L <= l && r <= R) {
return sum[o];
}
pushdown(o, l, r);
int mid = (l + r) >> 1;
int ans = 0;
if (L <= mid)ans += Sum(L, R, l, mid, ls[o]);
if (mid < R)ans += Sum(L, R, mid + 1, r, rs[o]);
return ans;
}
int Max(int L, int R, int l, int r, int o) {
if (L <= l && r <= R)return maxx[o];
pushdown(o, l, r);
int mid = (l + r) >> 1;
int MAX = -inf;
if (L <= mid)MAX = max(MAX, Max(L, R, l, mid, ls[o]));
if (mid < R)MAX = max(MAX, Max(L, R, mid + 1, r, rs[o]));
return MAX;
}
int Min(int L, int R, int l, int r, int o) {
if (L <= l && r <= R)return minn[o];
pushdown(o, l, r);
int mid = (l + r) >> 1;
int MIN = inf;
if (L <= mid)MIN = min(MIN, Min(L, R, l, mid, ls[o]));
if (mid < R)MIN = min(MIN, Min(L, R, mid + 1, r, rs[o]));
return MIN;
}
}t[22];
int main()
{
// ios::sync_with_stdio(0);
int r, c, m;
while (cin >> r >> c >> m) {
for (int i = 1; i <= r; i++)t[i].init();
while (m--) {
int opt = rd();
if (opt == 1) {
int X1 = rd(), Y1 = rd(), X2 = rd(), Y2 = rd(), v = rd();
for (int i = X1; i <= X2; i++)t[i].upd(t[i].rt, Y1, Y2, 1, c, 1, v);
}
else if (opt == 2) {
int X1 = rd(), Y1 = rd(), X2 = rd(), Y2 = rd(), v = rd();
for (int i = X1; i <= X2; i++)t[i].upd(t[i].rt, Y1, Y2, 1, c, 2, v);
}
else {
int ans = 0, ans1 = inf, ans2 = -inf;
int X1 = rd(), Y1 = rd(), X2 = rd(), Y2 = rd();
for (int i = X1; i <= X2; i++) {
ans += t[i].Sum(Y1, Y2, 1, c, t[i].rt);
ans1 = min(ans1, t[i].Min(Y1, Y2, 1, c, t[i].rt));
ans2 = max(ans2, t[i].Max(Y1, Y2, 1, c, t[i].rt));
}
printf("%d %d %d\n", ans, ans1, ans2);
}
}
}
return 0;
}
EPFL - Fighting
