传送门
文章目录
- 题意:
- 思路:
题意:
给你一个串,每次将区间都修改为某一个字母,问最终包含多少个FeiMaFeiMaFeiMa子序列。
思路:
首先暴力修改肯定是不行的,复杂度nqnqnq。
如果没有修改操作,有一个显然的dpdpdp方程,dp[i][j]dp[i][j]dp[i][j]表示到了第iii个,jjj对应FeiMaFeiMaFeiMa某个子序列的数量,考虑用线段树优化这个dpdpdp。
定义dp[u][l][r]dp[u][l][r]dp[u][l][r]表示到了线段树第uuu个子树包含FeiMaFeiMaFeiMa的[l,r][l,r][l,r]序列的个数,答案即为dp[1][0][4]dp[1][0][4]dp[1][0][4]。转移方程也比较明显了:dp[u][l][r]=dp[u<<1][l][r]+dp[u<<1∣1][l][r]+∑k=lr−1(dp[u<<1][l][k]+dp[u<<1∣1][k+1][r])dp[u][l][r]=dp[u<<1][l][r]+dp[u<<1|1][l][r]+\sum _{k=l} ^{r-1}(dp[u<<1][l][k]+dp[u<<1|1][k+1][r])dp[u][l][r]=dp[u<<1][l][r]+dp[u<<1∣1][l][r]+k=l∑r−1(dp[u<<1][l][k]+dp[u<<1∣1][k+1][r])
这个直接在pushuppushuppushup中维护即可。
对于区间修改,我们维护一个懒标记即可,修改的时候(如果是FeiMaFeiMaFeiMa的某个字母的话)直接让dp[u][x][x]=Len(u)dp[u][x][x]=Len(u)dp[u][x][x]=Len(u)即可。
// Problem: ★★飞马祝福语★★
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/16520/D
// Memory Limit: 524288 MB
// Time Limit: 6000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=100010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;int n,q;
char s[N];
string has="FeiMa";
struct Node {int l,r;int lazy;LL dp[5][5];
}tr[N<<2];int get(char ch) {if(ch=='F') return 0;else if(ch=='e') return 1;else if(ch=='i') return 2;else if(ch=='M') return 3;else if(ch=='a') return 4;else return 5;
}void calc(int u,int x) {for(int i=0;i<5;i++) for(int j=0;j<5;j++) tr[u].dp[i][j]=0;if(x<5) tr[u].dp[x][x]=(tr[u].r-tr[u].l+1)%mod;
}void pushup(int u) {for(int l=0;l<5;l++) {for(int r=l;r<5;r++) {tr[u].dp[l][r]=(tr[L].dp[l][r]+tr[R].dp[l][r])%mod;for(int k=l;k<=r-1;k++) tr[u].dp[l][r]+=tr[L].dp[l][k]*tr[R].dp[k+1][r]%mod,tr[u].dp[l][r]%=mod;}}
}void pushdown(int u) {if(tr[u].lazy==-1) return;tr[L].lazy=tr[R].lazy=tr[u].lazy;calc(L,tr[u].lazy); calc(R,tr[u].lazy);tr[u].lazy=-1;
}void build(int u,int l,int r) {tr[u]={l,r,-1};if(l==r) {calc(u,get(s[l]));return;}build(L,l,Mid); build(R,Mid+1,r);pushup(u);
}void change(int u,int l,int r,int x) {if(tr[u].l>=l&&tr[u].r<=r) {tr[u].lazy=x;calc(u,x);return;}pushdown(u);if(l<=Mid) change(L,l,r,x);if(r>Mid) change(R,l,r,x);pushup(u);
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d%d%s",&n,&q,s+1);build(1,1,n);while(q--) {int l,r;char ch;scanf("%d%d %c",&l,&r,&ch);change(1,l,r,get(ch));printf("%lld\n",tr[1].dp[0][4]%mod);}}return 0;
}
/**/