uva 219 - Department of Redundancy Department(dfs+剪枝)

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題目連結：uva 219 - Department of Redundancy Department

題目大意：給定一些關係，問哪一些關係是可以被替代的，如果可以被替代，給出替代的方案，一種即可。

解題思路：因為總共也就26個字母，所以用位元表示狀態。剪枝，每次將所有可選關係均考慮進去都無法滿足則是false。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;bool flag;int n, l[maxn], r[maxn];int cnt, ans[maxn], v[maxn];inline void cat (char* str, int x) {    bool sign = false;    int len = strlen(str);    for (int i = 0; i < len; i++) {        if (str[i] >= ‘A‘ && str[i] <= ‘Z‘) {            if (sign)                r[x] |= (1<<(str[i] - ‘A‘));            else                l[x] |= (1<<(str[i] - ‘A‘));        } else            sign = true;    }}inline void init () {    char str[maxn];    flag = true;    memset(l, 0, sizeof(l));    memset(r, 0, sizeof(r));    for (int i = 0; i < n; i++) {        scanf("%s", str);        cat(str, i);    }}inline void put (int k) {    printf("     FD %d is redundant using FDs:", k + 1);    for (int i = 0; ans[i] != -1; i++)        printf(" %d", ans[i]);    printf("\n");}bool check (int s, int k) {    int vis[maxn];    memset(vis, 0, sizeof(vis));    while (true) {        bool stop = true;        for (int i = 0; i < n; i++) {            if (vis[i] || v[i])                continue;            if ((s & l[i]) != l[i])                continue;            vis[i] = 1;            s |= r[i];            stop = false;        }        if (stop)            break;    }    return (s | r[k]) != s;}bool dfs (int d, int s, int k) {    ans[d] = -1;    if ((s & r[k]) == r[k])        return true;    if (check(s, k))        return false;    for (int i = 0; i < n; i++) {        if (v[i])            continue;        if ((s & l[i]) != l[i])            continue;        if ((s | r[i]) == s)            continue;        v[i] = 1;        if (dfs(d, s, k))            return true;        ans[d] = i + 1;        if (dfs(d+1, s | r[i], k))            return true;        v[i] = 0;    }    return false;}void solve () {    for (int i = 0; i < n; i++) {        memset(v, 0, sizeof(v));        v[i] = 1;        if (dfs(0, l[i], i)) {            put(i);            flag = false;        }    }}int main () {    int cas = 1;    while (scanf("%d", &n) == 1 && n) {        init();        printf("Set number %d\n", cas++);        solve();        if (flag)            printf("     No redundant FDs.\n");        printf("\n");    }    return 0;}