[USACO19JAN]Cow Poetry

题面

Solution:

这是一道很好的dp题。

一开始看不懂题面没有一点思路，看了好久题解才看懂题目...

(y[i]) 为第 (i) 个词结尾，(l[i]) 为第 (i) 个词长度。

设状态 (f[i][j]) 为长度为 (i) 的，以 (j) 结尾的一句诗的方案数，那么

[f[i][Y] = sum_{y[i]=Y}sum_{x=1}^{n}f[i-l[j]][x] ]

发现后面那一坨可以预处理，设

[g[i]=sum_{x=1}^nf[i][x] ]

(g[i]) 的意义是长度为 (i) 一句诗的方案数，显然可以无限背包 (O(nk)) 求。

[g[i]=sum_{j=1}^ng[i-l[j]] ]

那么

[f[i][Y]=sum_{y[j]=Y}g[i-l[j]] ]

对于每一个需要押的韵('A','B'...)答案就等于：

[Ans[i] = sum_{j=1}^nf[k][j]^{cnt[i]} ]

然后输出 (prod_limits{i=A}^ZAns[i])。

(Source:)

#include <set>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>

using namespace std;

#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define travle(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go
")

namespace io {
    static char buf[1<<21], *pos = buf, *end = buf;
    inline char getc()
    { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
    inline int rint() {
        register int x = 0, f = 1;register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline LL rLL() {
        register LL x = 0, f = 1; register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline void rstr(char *str) {
        while (isspace(*str = getc()));
        while (!isspace(*++str = getc()))
            if (*str == EOF) break;
        *str = '';
    }
    template<typename T> 
        inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
    template<typename T>
        inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }    
}
using namespace io;

const int N = 5e3 + 1, mod = 1e9 + 7;

int f[N][N], g[N], cnt[N], y[N], l[N], n, m, k;

LL qpow(LL a, LL b) {
    LL res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    file("Cow_Poetry");
#endif
    n = rint(), m = rint(), k = rint();
    for (register int i = 1; i <= n; ++ i) 
        l[i] = rint(), y[i] = rint();
    g[0] = 1;
    for (register int i = 1; i <= k; ++ i)
        for (register int j = 1; j <= n; ++ j) if (i - l[j] >= 0) 
            g[i] = (g[i] + g[i - l[j]]) % mod;
    for (register int i = 1; i <= k; ++ i)
        for (register int j = 1; j <= n; ++ j) if (i - l[j] >= 0)
            f[i][y[j]] = (f[i][y[j]] + g[i - l[j]]) % mod;
    char op[3];
    for (register int i = 1; i <= m; ++ i) {
        rstr(op);
        cnt[op[0] - 'A'] ++;
    }
    LL ans = 1;
    for (register int i = 0; i < 26; ++ i) if (cnt[i]) {
        LL res = 0;
        for (register int j = 1; j <= n; ++ j) if (f[k][j]) {
            res = (res + qpow(f[k][j], cnt[i]) % mod) % mod;
        }
        ans = ans * res % mod;
    }
    cout << ans << endl;
    return 0;
}