hdu 4689 Derangement（dp）

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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;}

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