Codeforces 1195E - OpenStreetMap

题目链接:[Codeforces 1195E]

典型二维RMQ,具体解法见我的上一篇博文BZOJ 1047 - 理想的正方形(题解)。[TODO: add links]

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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX_N = 3000 + 5;
int n, m, a, b;
LL g[MAX_N * MAX_N], x, y, z;
LL t[MAX_N][MAX_N];
LL getG(int u) {
if (g[u] != -1) return g[u];
else return g[u] = (getG(u - 1) * x % z + y % z) % z;
}
struct Node {
LL v, id;
friend bool operator == (const Node &nx, const Node &ny) {
return nx.id == ny.id && nx.v == ny.v;
}
};
// maintain MIN
struct GreaterQueue {
deque <Node> ori, q;
void push(LL x, int pos) {
ori.push_back({x, pos});
while (!q.empty() && q.back().v >= x) q.pop_back();
q.push_back({x, pos});
}
void pop() {
Node f = ori.front();
ori.pop_front();
if (f == q.front()) q.pop_front();
}
int front() {
return q.front().v;
}
void init() {
while (!q.empty()) q.pop_back();
while (!ori.empty()) ori.pop_back();
}
} qMin;
LL sum = 0;
int rMin[MAX_N][MAX_N], cMin[MAX_N][MAX_N];
int main() {
memset(g, -1, sizeof g);
scanf("%d%d%d%d", &n, &m, &a, &b);
scanf("%lld%lld%lld%lld", &g[0], &x, &y, &z);
g[0] %= z;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
t[i][j] = getG((i - 1) * m + j - 1);
// cout << t[i][j] << " ";
}
// cout << endl;
}
for (int i = 1; i <= n; ++i) {
qMin.init();
for (int j = 1; j <= m; ++j) {
if (j > b) qMin.pop();
qMin.push(t[i][j], j);
rMin[i][j] = qMin.front();
}
}
for (int j = 1; j <= m; ++j) {
qMin.init();
for (int i = 1; i <= n; ++i) {
if (i > a) qMin.pop();
qMin.push(rMin[i][j], i);
cMin[i][j] = qMin.front();
}
}
for (int i = a; i <= n; ++i) {
for (int j = b; j <= m; ++j) {
sum += cMin[i][j];
}
}
printf("%lld\n", sum);
return 0;
}