2017 多校4 Wavel Sequence
2017 多校4 Wavel Sequence
题意：
Formally, he defines a sequence \(a_1,a_2,...,a_n\) as ''wavel'' if and only if \(a_1<a_2>a_3<a_4>a_5<a_6\)...
Now given two sequences \(a_1,a_2,...,a_n\) and \(b_1,b_2,...,b_m\), Little Q wants to find two sequences \(f_1,f_2,...,f_k(1≤f_i≤n,f_i<f_{i+1})\) and \(g_1,g_2,...,g_k(1≤g_i≤m,g_i<g_i+1)\), where \(a_{f_i}=b_{g_i}\) always holds and sequence \(a_{f_1},a_{f_2},...,a_{f_k}\) is ''wavel''.
\(1<=n,m<=2000\)
\(1<=a_i,b_i<=2000\)
题解：
设\(f_{i,j,k}\)
表示仅考虑\(a[1..i]\)与\(b[1..j]\)，选择的两个子序列结尾分别是\(a_i\)和\(b_j\)，且上升下降状态是\(k\) 时的方案数，
则\(f_{i,j,k}=\sum f_{x,y,1-k}\)
，其中\(x<i,y<j\)
具体点说
定义\(f[i][j][0/1]\)为选择的两个子序列结尾分别是\(a_i\)和\(b_j\)，当前为下降/上升状态的方案数
则当\(a[i] = b[j]\)的时候有
\(f[i][j][0] = \sum f[x][y][1]\),其中\(x < i,y < j 且a[x] < a[i]\)
\(f[i][j][1] = \sum f[x][y][0] + 1\),其中\(x < i,y < j 且a[x] > a[i]\)
暴力枚举是O(n^4)的,可以用二维树状数组去优化两维变成\(O(n^{2}log{^2}n)\)
顺序枚举i，保证了第一维递增的，只需要用树状数组去维护第二维的下标和值
#include<bits/stdc++.h>
#define LL long long
#define P pair<int,int>
using namespace std;
const int N = 2e3 + 10;
const int mod = 998244353;
int read(){
int x = 0;
char c = getchar();
while(c < '0' || c > '9') c = getchar();
while(c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
return x;
}
int s[2][N][N];
int a[N],b[N];
int n,m;
void add(int &x,int y){
x += y;
if(x >= mod) x -= mod;
}
int lowbit(int x){
return x & (-x);
}
int sum(int o,int i,int j){
int ans = 0;
while(i){
int y = j;
while(y){
add(ans,s[o][i][y]);
y -= lowbit(y);
}
i -= lowbit(i);
}
return ans;
}
void update(int o,int i,int j,int val){
while(i <= n){
int y = j;
while(y <= 2000){
add(s[o][i][y],val);
y += lowbit(y);
}
i += lowbit(i);
}
}
int main(){
int T;
T = read();
while(T--){
n = read(),m = read();
for(int k = 0;k < 2;k++)
for(int i = 1;i <= n;i++)
for(int j = 1;j <= 2000;j++) s[k][i][j] = 0;
for(int i = 1;i <= n;i++) a[i] = read();
for(int i = 1;i <= m;i++) b[i] = read();
int ans = 0;
for(int i = 1;i <= m;i++){
for(int j = 1;j <= n;j++){
if(b[i] == a[j]){
int tmp1 = sum(1,j-1,a[j]-1),tmp2 = (mod + sum(0,j-1,2000)-sum(0,j-1,a[j]))%mod;
update(0,j,a[j],tmp1);/// 0 下降 1 上升
update(1,j,a[j],(tmp2 + 1)%mod);
add(ans,tmp1);
add(ans,tmp2);
add(ans,1);
}
}
}
printf("%d\n",ans);
}
return 0;
}
题解的\(O(n^2)\)的做法
用\(s[i][j][0/1]\)表示\(a\)和\(b\)分别在\(1\)$i$和$1$\(j\)的结尾的子序列的方案
那么\(dp[i][j][k] = s[i-1][j-1][1 - k] + k==1?1:0\)
\(i,j\)顺序枚举，遇到\(a[i] = b[j]\)的时候，前面可以顺便计算大于和小于它的方案，然后更新即可
#include<bits/stdc++.h>
#define LL long long
#define P pair<int,int>
using namespace std;
const int N = 2e3 + 10;
const int mod = 998244353;
int read(){
int x = 0;
char c = getchar();
while(c < '0' || c > '9') c = getchar();
while(c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
return x;
}
int s[2][N][N];
int dp[2][N][N];
int a[N],b[N];
int n,m;
void add(int &x,int y){
x += y;
if(x >= mod) x -= mod;
}
int main(){
int T;
T = read();
while(T--){
n = read(),m = read();
for(int i = 1;i <= n;i++) a[i] = read();
for(int i = 1;i <= m;i++) b[i] = read();
for(int k = 0;k < 2;k++)
for(int i = 1;i <= n;i++)
for(int j = 1;j <= m;j++) dp[k][i][j] = s[k][i][j] = 0;
int ans = 0;
for(int i = 1;i <= n;i++){
int tmp0 = 0,tmp1 = 0;///0 下降 1 上升
for(int j = 1;j <= m;j++){
if(a[i] == b[j]){
add(dp[0][i][j],tmp0);
add(dp[1][i][j],(tmp1+1)%mod);
add(ans,(dp[0][i][j]+dp[1][i][j])%mod);
}
else if(a[i] > b[j]){
add(tmp0, s[1][i-1][j]);
}else{
add(tmp1,s[0][i-1][j]);
}
}
for(int j = 1;j <= m;j++){
s[0][i][j] = s[0][i-1][j];
s[1][i][j] = s[1][i-1][j];
if(a[i] == b[j]){
add(s[0][i][j],dp[0][i][j]);
add(s[1][i][j],dp[1][i][j]);
}
}
}
printf("%d\n",ans);
}
return 0;
}
