【BZOJ】1566: [NOI2009]管道取珠

题解

假如我们非常熟练的看出来，平方和转有序对统计的套路的话，应该就不难了

我们只需要统计(wayA,wayB)生成的序列一样的有序对个数就行

可以用一个(n^3)的dp解决

(dp[i][j][k])表示选到第i个，第一个序列用j个上管道的球，第二个序列用了k的上管道的球，要求下一次操作两个球长得一样就可以了

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1024523;
int dp[2][505][505],N,M;
char up[505],down[505];
void update(int &x,int y) {
    x = x + y;
    if(x >= MOD) x -= MOD;
}
void Solve() {
    scanf("%d%d",&N,&M);
    scanf("%s",up + 1);scanf("%s",down + 1);
    up[N + 1] = 'C';down[M + 1] = 'D';
    int cur = 0;
    dp[0][0][0] = 1;
    for(int i = 0 ; i < N + M ; ++i) {
	int t = min(i,N);
	memset(dp[cur ^ 1],0,sizeof(dp[cur ^ 1]));
	for(int j = 0 ; j <= t ; ++j) {
	    for(int k = 0 ; k <= t ; ++k) {
		if(up[j + 1] == up[k + 1] && j != N && k != N) update(dp[cur ^ 1][j + 1][k + 1],dp[cur][j][k]);
		if(up[j + 1] == down[i - k + 1]) update(dp[cur ^ 1][j + 1][k],dp[cur][j][k]);
		if(down[i - j + 1] == up[k + 1]) update(dp[cur ^ 1][j][k + 1],dp[cur][j][k]);
		if(down[i - j + 1] == down[i - k + 1]) update(dp[cur ^ 1][j][k],dp[cur][j][k]);
	    }
	}
	cur ^= 1;
    }
    out(dp[cur][N][N]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}