Codeforces - 1195E - OpenStreetMap - 单调队列
作者:互联网
https://codeforc.es/contest/1195/problem/E
一个能运行但是会T的版本,因为本质上还是\(O(nmab)\)的算法。每次\(O(ab)\)初始化矩阵中的可能有用的点,然后\(O(n-a)\)往下推。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ERR(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {cerr << "\n";}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << "=" << a << ", ";
err(++it, args...);
}
#define ERR1(arg,n) { cerr<<""<<#arg<<"=\n "; for(int i=1;i<=n;i++) cerr<<arg[i]<<" "; cerr<<"\n"; }
#define ERR2(arg,n,m) { cerr<<""<<#arg<<"=\n"; for(int i=1;i<=n;i++) { cerr<<" "; for(int j=1;j<=m;j++)cerr<<arg[i][j]<<" "; cerr<<"\n"; } }
int n, m, a, b;
int x, y, z;
int g[3000 * 3000 + 5];
int h[3005][3005];
struct node {
int v, x;
node(int vv, int xx) {
v = vv;
x = xx;
}
};
int curx, cury;
deque<node> dq;
void calc(int x, int y) {
curx = x, cury = y;
while(!dq.empty())
dq.pop_back();
for(int i = 1; i <= a; i++) {
int minline = h[x + i - 1][y];
for(int j = 2; j <= b; j++) {
minline = min(minline, h[x + i - 1][y + j - 1]);
}
while(!dq.empty() && minline <= dq.back().v) {
dq.pop_back();
}
if(dq.empty() || minline > dq.back().v) {
dq.push_back(node(minline, x + i - 1));
}
}
}
void move_to_nextline() {
curx++;
if(dq.front().x < curx)
dq.pop_front();
int minline = h[curx + a - 1][cury];
for(int j = 2; j <= b; j++) {
minline = min(minline, h[curx + a - 1][cury + j - 1]);
}
while(!dq.empty() && minline <= dq.back().v) {
dq.pop_back();
}
if(dq.empty() || minline > dq.back().v) {
dq.push_back(node(minline, curx + a - 1));
}
}
ll ans = 0;
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
//freopen("Yinku.out", "w", stdout);
#endif // Yinku
while(~scanf("%d%d%d%d", &n, &m, &a, &b)) {
scanf("%d%d%d%d", &g[1], &x, &y, &z);
for(int i = 2; i <= n * m; i++)
g[i] = (1ll * g[i - 1] * x % z + y) % z;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
h[i][j] = g[(i - 1) * m + j];
//ERR2(h, n, m);
ans = 0;
for(int i = 1; i + a - 1 <= n; i++) {
for(int j = 1; j + b - 1 <= m; j++) {
calc(i, j);
ans += dq.front().v;
for(int di = 1; di <= a; di++) {
move_to_nextline();
}
//ERR(ans);
}
}
printf("%lld\n", ans);
}
}
标签:OpenStreetMap,args,int,d%,Codeforces,curx,back,1195E,dq 来源: https://www.cnblogs.com/Yinku/p/11205792.html