hdu 4689 Derangement（dp）

题目链接：hdu 4689 Derangement

解题思路

解法一：枚举二进制状态，为1的位置为满足+，为0的位置为满足-
解法二：dp，dp[i][j][k]表示到第i个位置，前面有j个-被满足，k个+未被满足

代码 - 解法一

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = (1<<20) + 5;
typedef unsigned long long ll;

int N, K, S, bit[maxn];
char str[25];

int bitcount(int s) { return s == 0 ? 0 : bitcount(s>>1) + (s&1); }

ll add(int s) {
    int c = 0;
    ll ret = 1;

    for (int i = 0; i < N; i++) {
        if (s&(1<<i)) {
            if (c == 0) return 0;
            ret *= c;
            c--;
        }
        if (str[i] == '+') c++;
    }
    return ret;
}

ll del(int s) {
    int c = 0;
    ll ret = 1;

    for (int i = N-1; i >= 0; i--) {
        if (s&(1<<i)) {
            if (c == 0) return 0;
            ret *= c;
            c--;
        }
        if (str[i] == '-') c++;
    }
    return ret;
}

ll solve () {
    N = strlen(str), K = 0, S = (1<<N)-1;
    for (int i = 0; i < N; i++) if (str[i] == '+') K++;

    ll ret = 0;
    for (int i = 0; i < (1<<N); i++) if (bit[i] == K) {
        ll l = add(i);
        if (l == 0) continue;
        ll r = del(i^S);
        ret += l * r;
    }
    return ret;
}

int main () {
    for (int i = 0; i < (1<<20); i++) bit[i] = bitcount(i);

    while (scanf("%s", str) == 1) {
        printf("%llu\n", solve());
    }
    return 0;
}

代码 - 解法二

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 25;
typedef unsigned long long ll;

char str[maxn];
ll dp[maxn][maxn][maxn];

ll solve () {
    int n = strlen(str), c = 0;
    for (int i = 0; i < n; i++) if (str[i] == '-') c++;

    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;

    for (int i = 0; i < n; i++) {
        if (str[i] == '+') {
            for (int x = 0; x < n; x++) {
                for (int y = 0; y < n; y++) {
                    if (dp[i][x][y] == 0) continue;

                    dp[i+1][x][y+1] += dp[i][x][y]; // 当前位置不变
                    if (i - x - y > 0) // 和1~i之间满足+的位置交换
                        dp[i+1][x][y+1] += dp[i][x][y] * (i-x-y);
                    dp[i+1][x][y] += dp[i][x][y] * y; // 和1~i之间不满足+的位置交换
                }
            }
        } else {
            for (int x = 0; x < n; x++) {
                for (int y = 0; y < n; y++) {
                    if (dp[i][x][y] == 0) continue;

                    if (i - x - y > 0) // 和1~i之间满足+的位置交换
                        dp[i+1][x+1][y] += dp[i][x][y] * (i-x-y);
                    if (y) // 和1~i之间不满足+的位置交换
                        dp[i+1][x+1][y-1] += dp[i][x][y] * y;
                }
            }
        }
    }
    return dp[n][c][0];
}

int main () {
    while (scanf("%s", str) == 1) {
        printf("%llu\n", solve());
    }
    return 0;
}