Bzoj1499: [NOI2005]瑰丽华尔兹

题面

Bzoj

Sol

暴力：(设f[i][j][k])表示到第(i)次倾斜，当前在((j, k))的滑动最大距离
然后(O(n*m*T))转移，(AC)了？？？

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(205);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, k, x, y, mp[_][_], T;
int s[_], t[_], d[_], dx[4] = {-1, 1}, dy[4] = {0, 0, -1, 1};
int f[2][_][_], ans;

int main(RG int argc, RG char* argv[]){
    n = Input(), m = Input(), x = Input(), y = Input(), k = Input();
    for(RG int i = 1; i <= n; ++i)
        for(RG int j = 1; j <= m; ++j){
            RG char op; scanf(" %c", &op);
            mp[i][j] = (op == 'x');
        }
    for(RG int i = 1; i <= k; ++i)
        s[i] = Input(), t[i] = Input(), d[i] = Input() - 1;
    Fill(f, -127), f[0][x][y] = 0;
    for(RG int T = 1; T <= k; ++T){
        RG int nxt = T & 1, lst = nxt ^ 1;
        for(RG int i = 1; i <= n; ++i)
            for(RG int j = 1; j <= m; ++j)
                f[nxt][i][j] = f[lst][i][j];
        for(RG int i = 1; i <= n; ++i)
            for(RG int j = 1; j <= m; ++j)
                for(RG int p = 0; p <= t[T] - s[T] + 1; ++p){
                    RG int xx = i + p * dx[d[T]], yy = j + p * dy[d[T]];
                    if(xx < 1 || yy < 1 || xx > n || yy > m || mp[xx][yy]) break;
                    f[nxt][xx][yy] = max(f[nxt][xx][yy], f[lst][i][j] + p);
                }
    }
    for(RG int i = 1; i <= n; ++i)
        for(RG int j = 1; j <= m; ++j)
            ans = max(ans, f[k & 1][i][j]);
    printf("%d
", ans);
    return 0;
}

然后每次要么是一行一行转移，要么一列一列转移
分开写发现是可以单调队列优化的

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(205);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, k, x, y, mp[_][_];
int s[_], t[_], d[_], dx[4] = {-1, 1}, dy[4] = {0, 0, -1, 1};
int f[_][_], ans;
struct Queue{
	int val, p;
} Q[_];

IL void Solve(RG int X, RG int Y, RG int T, RG int D){
	RG int l = 0, r = -1;
	for(RG int i = 0; ; ++i){
		if(mp[X][Y]) l = 0, r = -1;
		else{
			while(l <= r && f[X][Y] - i > Q[r].val) --r;
			Q[++r] = (Queue){f[X][Y] - i, i};
			while(l <= r && i - Q[l].p > T) ++l;
			f[X][Y] = Q[l].val + i;
		}
		X += dx[D], Y += dy[D];
		if(X < 1 || Y < 1 || X > n || Y > m) break;
	}
}

int main(RG int argc, RG char* argv[]){
	n = Input(), m = Input(), x = Input(), y = Input(), k = Input();
	for(RG int i = 1; i <= n; ++i)
		for(RG int j = 1; j <= m; ++j){
			RG char op; scanf(" %c", &op);
			mp[i][j] = (op == 'x');
		}
	for(RG int i = 1; i <= k; ++i)
		s[i] = Input(), t[i] = Input(), d[i] = Input() - 1;
	Fill(f, -127), f[x][y] = 0;
	for(RG int T = 1; T <= k; ++T){
		RG int len = t[T] - s[T] + 1;
		if(d[T] == 0)
			for(RG int i = 1; i <= m; ++i) Solve(n, i, len, 0);
		else if(d[T] == 1)
			for(RG int i = 1; i <= m; ++i) Solve(1, i, len, 1);
		else if(d[T] == 2)
			for(RG int i = 1; i <= n; ++i) Solve(i, m, len, 2);
		else
			for(RG int i = 1; i <= n; ++i) Solve(i, 1, len, 3);
	}
	for(RG int i = 1; i <= n; ++i)
		for(RG int j = 1; j <= m; ++j)
			ans = max(ans, f[i][j]);
	printf("%d
", ans);
    return 0;
}