hihocoder 1159 扑克牌(组合DP)

来源：互联网发布：美工与ui的区别编辑：程序博客网时间：2024/05/16 11:03

先预处理出每种牌的张数，对于花色只要考虑一下排列组合即可

那么设dp[i][j],表示用到第i张牌，相邻位置为相同花色有j个，然后枚举当前的牌要分成几堆，几堆插入到相同花色位置，剩下插入到不同花色位置，这样去转移即可

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long ll;const char card[15] = {"0A23456789TJQK"};int to[300], to2[300];int t, n, cnt[14][4];char s[3];ll C[55][55], dp[14][55];int main() {    for (int i = 0; i <= 50; i++) {        C[i][0] = C[i][i] = 1;        for (int j = 1; j < i; j++)            C[i][j] = C[i - 1][j - 1] + C[i - 1][j];    }    for (int i = 1; i <= 13; i++) to[card[i]] = i;    to2['S'] = 0; to2['H'] = 1; to2['D'] = 2; to2['C'] = 3;    scanf("%d", &t);    int cas = 0;    while (t--) {        scanf("%d", &n);        memset(cnt, 0, sizeof(cnt));        ll ans = 1;        while (n--) {            scanf("%s", s);            cnt[to[s[0]]][to2[s[1]]]++;        }        for (int i = 1; i <= 13; i++) {            int sum = 0;            for (int j = 0; j < 4; j++) sum += cnt[i][j];            for (int j = 0; j < 4; j++) ans *= C[sum][cnt[i][j]], sum -= cnt[i][j];        }        int tot = 0;        memset(dp, 0, sizeof(dp));        dp[0][0] = 1;        for (int i = 1; i <= 13; i++) {            int sum = 0;            for (int j = 0; j < 4; j++) sum += cnt[i][j];            if (sum == 0) {                for (int j = 0; j <= max(0, tot - 1); j++)                    dp[i][j] = dp[i - 1][j];                continue;            }            for (int j = 0; j <= max(0, tot - 1); j++) {                for (int k = 1; k <= sum; k++) {                    for (int x = 0; x <= k; x++) {                        if (j + sum - k - x < 0) continue;                        dp[i][j + sum - k - x] += dp[i - 1][j] * C[sum - 1][k - 1] * C[j][x] * C[tot + 1 - j][k - x];                    }                }            }            tot += sum;        }        printf("Case #%d: %llu\n", ++cas, dp[13][0] * ans);    }    return 0;}

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