HDU 4924 Football Manager(状压DP)

题目连接 ： http://acm.hdu.edu.cn/showproblem.php?pid=4924

题意 ： n(<=20)个人选出11个人作为首发组成a-b-c阵容，每个人都有自己擅长的位置，并且每个人和其他人会有单向的厌恶和喜欢关系，每个人对于自己擅长的位置都有两个值CA和PA，有喜欢或者厌恶关系的两个人在一起也会影响整个首发的CA总值，要求选出一套阵容使得CA最大，或者CA一样的情况下PA最大。

思路 ： 状压搞，dp[s]s的二进制表示20个人中选了那几个人，然后规定选进来的顺序就是那个人所担当的位置，这样就可以避免知道状态的情况下去算最优而外加的复杂度了。

PS ：不过这道题目时限似乎卡得很紧，注意用lowbit()优化+其它一些优化，另外注意一些输入信息，关系中a和b有可能重复等细节就可以AC了。

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

using namespace std;
map<int,int> mp;
const int INF = 0x3f3f3f3f;
typedef pair<int,int> PII;
const int MAXN = 1<<20;
int lowbit(int x) {return x&-x;}

int num[MAXN] = {0};
int has[MAXN] = {0};
int vis[MAXN] = {0};
int Has[MAXN] = {0};
int ans[MAXN];
int abl[55][10];
PII val[55][10];
PII dp[MAXN];
int state[22];
int G[22][22];
int n;
int tot;
int E;
void BUG(int a[], int n) {
    for (int i = 1; i <= n; i++) {
        printf("%d%c", a[i], " \n"[i == n]);
    }
}

void init() {
    num[0] = 0; vis[0] = 1;
    tot = 1; E = 0;
    for (int i = 1; i < MAXN; i++) {
        if (!vis[i-lowbit(i)] || num[i-lowbit(i)] == 11)continue;
        num[i] = num[i-lowbit(i)] + 1; has[i] = tot; Has[tot] = i; vis[i] = 1;
        if (num[i] == 11) {
            ans[++ E] = tot;
        }
        tot ++;
    }
    for (int i = 0; i < 20; i++) {
        vis[1<<i] = i + 1;
    }
    //printf("tot = %d\n", tot);
    return ;
}

PII update(PII a, PII b) {
    return a < b ? b : a;
    /*
    if (a.first > b.first || a.first == b.first && a.second > b.second) {
        return a;
    }else {
        return b;
    }
    */
}
int c1, c2, c3, c4;
void add(int e, char s[], int C, int P) {
    if (e > n) while (true);
    if (s[0] == 'G') {
        val[e][1] = make_pair(C, P);
        abl[e][1] = 1;
        ++ c1;
    }else if (s[0] == 'D') {
        val[e][2] = make_pair(C, P);
        abl[e][2] = 1;
        ++ c2;
    }else if (s[0] == 'M') {
        val[e][3] = make_pair(C, P);
        abl[e][3] = 1;
        ++ c3;
    }else {
        val[e][4] = make_pair(C, P);
        abl[e][4] = 1;
        ++ c4;
    }
}
bool used[MAXN];
void solve() {
    memset(used, false, sizeof(used));
    PII ret = make_pair(-INF, -INF);
    dp[0] = make_pair(0, 0); used[0] = 1;
    for (int ti = 1; ti < tot; ti++) {
        int cnt = num[Has[ti]];
        int i = Has[ti];
        dp[ti] = make_pair(-INF, -INF);
        for (int j = i; j > 0; j -= lowbit(j)) {
            int nex = vis[lowbit(j)];
            int ts = has[i-lowbit(j)];
            if (!used[ts]) continue;
            if (!abl[nex][state[cnt]]) continue;
            dp[ti] = update(dp[ti], make_pair(dp[ts].first+val[nex][state[cnt]].first, dp[ts].second+val[nex][state[cnt]].second));
            used[ti] = 1;
        }
        if (cnt == 11 && used[ti]) {
            int tmp = 0;
            for (int j = i; j > 0; j -= lowbit(j)) {
                int now = vis[lowbit(j)];
                for (int k = j; k > 0; k -= lowbit(k)) {
                    int nex = vis[lowbit(k)];
                    if (now != nex)tmp += G[now][nex] + G[nex][now];
                    else tmp += G[now][nex];
                }
            }
            ret = update(ret, make_pair(dp[ti].first+tmp, dp[ti].second));
        }
    }
    if (ret.first == -INF) {
        printf("Poor Manager!\n");
    }else {
        printf("%d %d\n", ret.first, ret.second);
    }
    return ;
}

int main() {
    init();
    int T;
    scanf("%d", &T);
    for (int cas = 1; cas <= T; cas++) {
        scanf("%d", &n);
        memset(abl, 0, sizeof(abl));
        memset(G, 0, sizeof(G));
        c1 = 0, c2 = 0, c3 = 0, c4 = 0;
        mp.clear();
        for (int i = 1; i <= n; i++) {
            int e, m;
            scanf("%d%d", &e, &m);
            mp[e] = i;
            char ts[5];
            int tc, tp;
            for (int j = 1; j <= m; j++) {
                scanf("%s %d%d", ts, &tc, &tp);
                add(i, ts, tc, tp);
            }
        }
        int Q;
        scanf("%d", &Q);
        while (Q--) {
            int a, b, c;
            char ts[11];
            scanf("%d%d%s%d", &a, &b, ts, &c);
            a = mp[a]; b = mp[b];
            if (ts[0] == 'L') {
                G[a][b] = c;
            }else {
                G[a][b] = -c;
            }
        }
        int nd, nm, ns;
        scanf("%d-%d-%d", &nd, &nm, &ns);
        if (n < 11 || c1 < 1 || c2 < nd || c3 < nm || c4 < ns) {
            printf("Poor Manager!\n"); continue;
        }
        state[1] = 1;
        for (int i = 2; i <= nd+1; i++) state[i] = 2;
        for (int i = 2+nd; i <= nd+1+nm; i++) state[i] = 3;
        for (int i = 2+nd+nm; i <= 11; i++) state[i] = 4;
        //for (int i = 1; i <= 11; i++) printf("%d%c", state[i], " \n"[i == 11]);
        //BUG(state, 11);
        solve();
    }
    return 0;
}

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