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。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <map>
5
6 using namespace std;
7 map<int,int> mp;
8 const int INF = 0x3f3f3f3f;
9 typedef pair<int,int> PII;
10 const int MAXN = 1<<20;
11 int lowbit(int x) {return x&-x;}
12
13 int num[MAXN] = {0};
14 int has[MAXN] = {0};
15 int vis[MAXN] = {0};
16 int Has[MAXN] = {0};
17 int ans[MAXN];
18 int abl[55][10];
19 PII val[55][10];
20 PII dp[MAXN];
21 int state[22];
22 int G[22][22];
23 int n;
24 int tot;
25 int E;
26 void BUG(int a[], int n) {
27 for (int i = 1; i <= n; i++) {
28 printf("%d%c", a[i], " \n"[i == n]);
29 }
30 }
31
32 void init() {
33 num[0] = 0; vis[0] = 1;
34 tot = 1; E = 0;
35 for (int i = 1; i < MAXN; i++) {
36 if (!vis[i-lowbit(i)] || num[i-lowbit(i)] == 11)continue;
37 num[i] = num[i-lowbit(i)] + 1; has[i] = tot; Has[tot] = i; vis[i] = 1;
38 if (num[i] == 11) {
39 ans[++ E] = tot;
40 }
41 tot ++;
42 }
43 for (int i = 0; i < 20; i++) {
44 vis[1<<i] = i + 1;
45 }
46 //printf("tot = %d\n", tot);
47 return ;
48 }
49
50 PII update(PII a, PII b) {
51 return a < b ? b : a;
52 /*
53 if (a.first > b.first || a.first == b.first && a.second > b.second) {
54 return a;
55 }else {
56 return b;
57 }
58 */
59 }
60 int c1, c2, c3, c4;
61 void add(int e, char s[], int C, int P) {
62 if (e > n) while (true);
63 if (s[0] == ‘G‘) {
64 val[e][1] = make_pair(C, P);
65 abl[e][1] = 1;
66 ++ c1;
67 }else if (s[0] == ‘D‘) {
68 val[e][2] = make_pair(C, P);
69 abl[e][2] = 1;
70 ++ c2;
71 }else if (s[0] == ‘M‘) {
72 val[e][3] = make_pair(C, P);
73 abl[e][3] = 1;
74 ++ c3;
75 }else {
76 val[e][4] = make_pair(C, P);
77 abl[e][4] = 1;
78 ++ c4;
79 }
80 }
81 bool used[MAXN];
82 void solve() {
83 memset(used, false, sizeof(used));
84 PII ret = make_pair(-INF, -INF);
85 dp[0] = make_pair(0, 0); used[0] = 1;
86 for (int ti = 1; ti < tot; ti++) {
87 int cnt = num[Has[ti]];
88 int i = Has[ti];
89 dp[ti] = make_pair(-INF, -INF);
90 for (int j = i; j > 0; j -= lowbit(j)) {
91 int nex = vis[lowbit(j)];
92 int ts = has[i-lowbit(j)];
93 if (!used[ts]) continue;
94 if (!abl[nex][state[cnt]]) continue;
95 dp[ti] = update(dp[ti], make_pair(dp[ts].first+val[nex][state[cnt]].first, dp[ts].second+val[nex][state[cnt]].second));
96 used[ti] = 1;
97 }
98 if (cnt == 11 && used[ti]) {
99 int tmp = 0;
100 for (int j = i; j > 0; j -= lowbit(j)) {
101 int now = vis[lowbit(j)];
102 for (int k = j; k > 0; k -= lowbit(k)) {
103 int nex = vis[lowbit(k)];
104 if (now != nex)tmp += G[now][nex] + G[nex][now];
105 else tmp += G[now][nex];
106 }
107 }
108 ret = update(ret, make_pair(dp[ti].first+tmp, dp[ti].second));
109 }
110 }
111 if (ret.first == -INF) {
112 printf("Poor Manager!\n");
113 }else {
114 printf("%d %d\n", ret.first, ret.second);
115 }
116 return ;
117 }
118
119 int main() {
120 init();
121 int T;
122 scanf("%d", &T);
123 for (int cas = 1; cas <= T; cas++) {
124 scanf("%d", &n);
125 memset(abl, 0, sizeof(abl));
126 memset(G, 0, sizeof(G));
127 c1 = 0, c2 = 0, c3 = 0, c4 = 0;
128 mp.clear();
129 for (int i = 1; i <= n; i++) {
130 int e, m;
131 scanf("%d%d", &e, &m);
132 mp[e] = i;
133 char ts[5];
134 int tc, tp;
135 for (int j = 1; j <= m; j++) {
136 scanf("%s %d%d", ts, &tc, &tp);
137 add(i, ts, tc, tp);
138 }
139 }
140 int Q;
141 scanf("%d", &Q);
142 while (Q--) {
143 int a, b, c;
144 char ts[11];
145 scanf("%d%d%s%d", &a, &b, ts, &c);
146 a = mp[a]; b = mp[b];
147 if (ts[0] == ‘L‘) {
148 G[a][b] = c;
149 }else {
150 G[a][b] = -c;
151 }
152 }
153 int nd, nm, ns;
154 scanf("%d-%d-%d", &nd, &nm, &ns);
155 if (n < 11 || c1 < 1 || c2 < nd || c3 < nm || c4 < ns) {
156 printf("Poor Manager!\n"); continue;
157 }
158 state[1] = 1;
159 for (int i = 2; i <= nd+1; i++) state[i] = 2;
160 for (int i = 2+nd; i <= nd+1+nm; i++) state[i] = 3;
161 for (int i = 2+nd+nm; i <= 11; i++) state[i] = 4;
162 //for (int i = 1; i <= 11; i++) printf("%d%c", state[i], " \n"[i == 11]);
163 //BUG(state, 11);
164 solve();
165 }
166 return 0;
167 }
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HDU 4924 Football Manager(状压DP),布布扣,bubuko.com
时间： 2024-08-06 11:57:32