Codeforces Round #439 (Div. 2) E. The Untended Antiquity (hash+数状数组)
2017-10-07 12:50
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这个题,做出来的人很多,我感觉是数据不够强,我看了很多人的代码直接暴力也能过了,直接暴力如果数据够强的话肯定是时间超限,边缘数据不够强。如果和上次一样估计很多人的E会GG。我看到一位OIdalao的代码,认为这个是正确的解法,对每一道围墙进行hash处理,然后用二维的树状数组来解决这个问题。感觉博主已经写得简单易懂了。长了姿势
#include <bits/stdc++.h>
using namespace std;
map<pair<pair<int, int>, pair<int, int> >, int>g;
int n, m, q;
long long c[2520][2520];
void R(int x, int y, long long z) {
for (int i = x; i <= n; i += i & -i) {
for (int j = y; j <= m; j += j & -j) {
c[i][j] += z;
}
}
}
long long G(int x, int y) {
long long re = 0;
for (int i = x; i > 0; i -= i & -i) {
for (int j = y; j > 0; j -= j & -j) {
re += c[i][j];
}
}
return re;
}
int rd() {
return rand() << 15 | rand();
}
int main() {
scanf("%d%d%d", &n, &m, &q);
srand(time(0));
for (int i = 0; i < 011; i++) {
srand(rd());
}
for (int i = 0; i < q; i++) {
int o, xa, xb, ya, yb;
scanf("%d%d%d%d%d", &o, &xa, &ya, &xb, &yb);
if (o == 1) {
int u = rd();
g[make_pair(make_pair(xa, ya), make_pair(xb, yb))] = u;
R(xa, ya, u);
R(xa, yb + 1, -u);
R(xb + 1, ya, -u);
R(xb + 1, yb + 1, u);
} else if (o == 2) {
int u = g[make_pair(make_pair(xa, ya), make_pair(xb, yb))];
R(xa, ya, -u);
R(xa, yb + 1, u);
R(xb + 1, ya, u);
R(xb + 1, yb + 1, -u);
} else {
long long va = G(xa, ya);
long long vb = G(xb, yb);
if (va == vb) {
printf("Yes\n");
} else {
printf("No\n");
}
}
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
map<pair<pair<int, int>, pair<int, int> >, int>g;
int n, m, q;
long long c[2520][2520];
void R(int x, int y, long long z) {
for (int i = x; i <= n; i += i & -i) {
for (int j = y; j <= m; j += j & -j) {
c[i][j] += z;
}
}
}
long long G(int x, int y) {
long long re = 0;
for (int i = x; i > 0; i -= i & -i) {
for (int j = y; j > 0; j -= j & -j) {
re += c[i][j];
}
}
return re;
}
int rd() {
return rand() << 15 | rand();
}
int main() {
scanf("%d%d%d", &n, &m, &q);
srand(time(0));
for (int i = 0; i < 011; i++) {
srand(rd());
}
for (int i = 0; i < q; i++) {
int o, xa, xb, ya, yb;
scanf("%d%d%d%d%d", &o, &xa, &ya, &xb, &yb);
if (o == 1) {
int u = rd();
g[make_pair(make_pair(xa, ya), make_pair(xb, yb))] = u;
R(xa, ya, u);
R(xa, yb + 1, -u);
R(xb + 1, ya, -u);
R(xb + 1, yb + 1, u);
} else if (o == 2) {
int u = g[make_pair(make_pair(xa, ya), make_pair(xb, yb))];
R(xa, ya, -u);
R(xa, yb + 1, u);
R(xb + 1, ya, u);
R(xb + 1, yb + 1, -u);
} else {
long long va = G(xa, ya);
long long vb = G(xb, yb);
if (va == vb) {
printf("Yes\n");
} else {
printf("No\n");
}
}
}
return 0;
}
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