The 2018 ICPC Asia-East Continent Final E（dp）

可以发现两个串合法的充要条件：

令P=−1,V=1P=-1,V = 1P=−1,V=1，条件是两个串的最小前缀和之和≥−1\ge-1≥−1
注：这里最小前缀和只能是以PPP结尾，或者整个串没有PPP
最小前缀和取到整个串的时候比较特殊
然后就得到了一个显然的做法
f[i][j][k]f[i][j][k]f[i][j][k]表示考虑串的前iii位，当前前缀和为jjj，最小前缀和为kkk的方案数
直接转移即可

这个dpdpdp几乎是无法优化的
正难则返
f[i][j]f[i][j]f[i][j]表示考虑串的后iii位，从当前位开始的最小前缀和是jjj
再合并
然后直接做就行了。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x],y = e[i].y;i;i = e[i].n,y = e[i].y)
#define P pair<int,int>
#define Pil pair<int,ll>
#define Pli pair<ll,int>
#define Pll pair<ll,ll>
#define pb push_back
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
#define fr first
#define se second
int rd()
{int num = 0;char c = getchar();bool flag = true;while(c < '0'||c > '9') {if(c == '-') flag = false;c = getchar();}while(c >= '0' && c <= '9') num = num*10+c-48,c = getchar();if(flag) return num;else return -num;
}
const int p = 998244353;
int trf[2][2][10100],trg[5010][2][10100];
int *f[2][2],*g[5010][2];
int p1,p2,n,m,det;
char s[5010],t[5010];
bool ck[5010];
inline int mul(int a,int b){return 1ll*a*b%p;}
inline void add(int &a,int b){a += b;if(a>=p) a-=p;}
inline int calc(int a,int b){return (a+=b)>=p?a-=p:a;}
inline int max(int a,int b){return a>b?a:b;}
int dp(int n,char *s,int *g[2][2])
{int f = 1;g[0][1][0] = 1;repp(i,n,1){rep(j,i-n,n-i)if(g[f^1][0][j] || g[f^1][1][j]){if(s[i] != 'P') add(g[f][0][j+1],g[f^1][0][j]),add(g[f][1][j+1],g[f^1][1][j]);if(s[i] != 'V') if(j>0) add(g[f][0][-1],calc(g[f^1][0][j],g[f^1][1][j]));else add(g[f][0][j-1],g[f^1][0][j]),add(g[f][1][j-1],g[f^1][1][j]); g[f^1][0][j] = g[f^1][1][j] = 0;}f ^= 1;}return f^1;
}
int main()
{rep(i,0,1) rep(j,0,1) f[i][j] = trf[i][j]+5010;rep(i,0,5001) rep(j,0,1)g[i][j] = trg[i][j]+5010;det = max(n,m);scanf("%s",s+1);scanf("%s",t+1);n = strlen(s+1);m = strlen(t+1);p1 = dp(n,s,f);int ans = 0;g[m+1][1][0] = 1;repp(i,m,1)rep(j,i-m,m-i){if(t[i] != 'P') add(g[i][0][j+1],g[i+1][0][j]),add(g[i][1][j+1],g[i+1][1][j]);if(t[i] != 'V') if(j>0) add(g[i][0][-1],calc(g[i+1][0][j],g[i+1][1][j]));else add(g[i][0][j-1],g[i+1][0][j]),add(g[i][1][j-1],g[i+1][1][j]); }rep(i,1,m) if(t[i] == 'P') break;else ck[i] = true;rep(i,1,n+1) repp(j,max(n,m)+1,-max(m,n)-1) add(g[i][0][j],g[i][0][j+1]),add(g[i][1][j],g[i][1][j+1]);rep(i,-n,n)//f[1/0][i][1/-]{if(i<0&&!ck[abs(i)]) continue;add(ans,mul(f[p1][0][i],calc(g[max(1-i,1)][0][i<=0?-1:-1-i],g[max(1-i,1)][1][i<=0?0:-i])));add(ans,mul(f[p1][1][i],calc(g[max(1-i,1)][0][i<=0?0:-i],g[max(1-i,1)][1][i<=0?0:-i])));}printf("%d\n",ans);return 0;
}

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