一定要注意审题啊，题目说的是选出做少的英雄打败其余处在任何模式下的英雄。
共有Sigma(num of model)个方案，每个方案有Sigma(num of model)+n个决策。
挺不错的一道精确覆盖的题目，最近发现精确覆盖很有意思。
1 /* 3957 */
2 #include <iostream>
3 #include <string>
4 #include <map>
5 #include <queue>
6 #include <set>
7 #include <stack>
8 #include <vector>
9 #include <deque>
10 #include <algorithm>
11 #include <cstdio>
12 #include <cmath>
13 #include <ctime>
14 #include <cstring>
15 #include <climits>
16 #include <cctype>
17 #include <cassert>
18 #include <functional>
19 #include <iterator>
20 #include <iomanip>
21 using namespace std;
22 //#pragma comment(linker,"/STACK:102400000,1024000")
23
24 #define sti set<int>
25 #define stpii set<pair<int, int> >
26 #define mpii map<int,int>
27 #define vi vector<int>
28 #define pii pair<int,int>
29 #define vpii vector<pair<int,int> >
30 #define rep(i, a, n) for (int i=a;i<n;++i)
31 #define per(i, a, n) for (int i=n-1;i>=a;--i)
32 #define clr clear
33 #define pb push_back
34 #define mp make_pair
35 #define fir first
36 #define sec second
37 #define all(x) (x).begin(),(x).end()
38 #define SZ(x) ((int)(x).size())
39 #define lson l, mid, rt<<1
40 #define rson mid+1, r, rt<<1|1
41
42 typedef struct {
43 static const int maxc = 30*3+5;
44 static const int maxr = 60;
45 static const int maxn = 60*60+5;
46
47 int nmodel;
48 int n, sz;
49 int S[maxc];
50
51 int col[maxn];
52 int L[maxn], R[maxn], U[maxn], D[maxn];
53 bool visit[maxc];
54
55 int ansd;
56
57 void init(int nmodel_, int n_) {
58 ansd = INT_MAX;
59 n = n_;
60 nmodel = nmodel_;
61
62 rep(i, 0, n+1) {
63 L[i] = i - 1;
64 R[i] = i + 1;
65 U[i] = i;
66 D[i] = i;
67 col[i] = i;
68 }
69 L[0] = n;
70 R[n] = 0;
71
72 sz = n + 1;
73 memset(S, 0, sizeof(S));
74 }
75
76 void addRow(vi columns) {
77 int first = sz;
78 int size = SZ(columns);
79
80 rep(i, 0, size) {
81 int c = columns[i];
82
83 L[sz] = sz - 1;
84 R[sz] = sz + 1;
85
86 D[sz] = c;
87 U[sz] = U[c];
88 D[U[c]] = sz;
89 U[c] = sz;
90
91 col[sz] = c;
92
93 ++S[c];
94 ++sz;
95 }
96
97 L[first] = sz - 1;
98 R[sz - 1] = first;
99 }
100
101 void remove(int c) {
102 L[R[c]] = L[c];
103 R[L[c]] = R[c];
104 for (int i=D[c]; i!=c; i=D[i]) {
105 for (int j=R[i]; j!=i; j=R[j]) {
106 U[D[j]] = U[j];
107 D[U[j]] = D[j];
108 --S[col[j]];
109 }
110 }
111 }
112
113 void restore(int c) {
114 L[R[c]] = c;
115 R[L[c]] = c;
116 for (int i=D[c]; i!=c; i=D[i]) {
117 for (int j=R[i]; j!=i; j=R[j]) {
118 U[D[j]] = j;
119 D[U[j]] = j;
120 ++S[col[j]];
121 }
122 }
123 }
124
125 void remove_col(int c) {
126 for (int i=D[c]; i!=c; i=D[i]) {
127 L[R[i]] = L[i];
128 R[L[i]] = R[i];
129 }
130 }
131
132 void restore_col(int c) {
133 for (int i=D[c]; i!=c; i=D[i]) {
134 L[R[i]] = i;
135 R[L[i]] = i;
136 }
137 }
138
139 int H() {
140 // return 0;
141 int ret = 0;
142
143 memset(visit, false, sizeof(visit));
144 for (int c=R[0]; c!=0&&c<=nmodel; c=R[c]) {
145 if (visit[c])
146 continue;
147 ++ret;
148 visit[c] = true;
149 for (int i=D[c]; i!=c; i=D[i]) {
150 for (int j=R[i]; j!=i; j=R[j])
151 visit[col[j]] = true;
152 }
153 }
154
155 return ret;
156 }
157
158 void dfs(int d) {
159 if (R[0]==0 || R[0]>nmodel) {
160 ansd = min(ansd, d);
161 return ;
162 }
163
164 int delta = H();
165
166 if (d+delta >= ansd)
167 return ;
168
169 int c = R[0];
170 for (int i=R[0]; i!=0&&i<=nmodel; i=R[i]) {
171 if (S[i] < S[c])
172 c = i;
173 }
174
175 for (int i=D[c]; i!=c; i=D[i]) {
176 remove_col(i);
177 for (int j=R[i]; j!=i; j=R[j]) {
178 if (col[j] <= nmodel)
179 remove_col(j);
180 }
181 for (int j=R[i]; j!=i; j=R[j]) {
182 if (col[j] > nmodel)
183 remove(col[j]);
184 }
185 dfs(d + 1);
186 for (int j=L[i]; j!=i; j=L[j]) {
187 if (col[j] > nmodel)
188 restore(col[j]);
189 }
190 for (int j=L[i]; j!=i; j=L[j]) {
191 if (col[j] <= nmodel)
192 restore_col(j);
193 }
194 restore_col(i);
195 }
196 }
197
198 void solve() {
199 ansd = INT_MAX;
200 dfs(0);
201 }
202
203 } DLX;
204
205 typedef struct {
206 int m, k[2];
207 pii p[2][11];
208 } hero_t;
209
210 DLX solver;
211 int ID[30][2];
212 int nmodel, n;
213 hero_t hero[30];
214
215 void solve() {
216 solver.init(nmodel, n+nmodel);
217
218 rep(i, 0, n) {
219 rep(j, 0, hero[i].m) {
220 // each model has a plan
221 vi columns;
222 // cover i-th hero
223 columns.pb(nmodel+i+1);
224 // cover 0-th plan
225 columns.pb(ID[i][0]);
226 // may cover 1-th plan
227 if (hero[i].m == 2)
228 columns.pb(ID[i][1]);
229 rep(k, 0, hero[i].k[j]) {
230 int role_id = hero[i].p[j][k].fir;
231 int model_id = hero[i].p[j][k].sec;
232 columns.pb(ID[role_id][model_id]);
233 }
234
235 solver.addRow(columns);
236 }
237 }
238
239 solver.solve();
240
241 printf("%d\n", solver.ansd);
242 }
243
244 int main() {
245 ios::sync_with_stdio(false);
246 #ifndef ONLINE_JUDGE
247 freopen("data.in", "r", stdin);
248 freopen("data.out", "w", stdout);
249 #endif
250
251 int t;
252
253 scanf("%d", &t);
254 rep(tt, 1, t+1) {
255 scanf("%d", &n);
256 nmodel = 0;
257 rep(i, 0, n) {
258 scanf("%d", &hero[i].m);
259 rep(j, 0, hero[i].m) {
260 scanf("%d", &hero[i].k[j]);
261 rep(k, 0, hero[i].k[j])
262 scanf("%d %d", &hero[i].p[j][k].fir, &hero[i].p[j][k].sec);
263 ID[i][j] = ++nmodel;
264 }
265 }
266 printf("Case %d: ", tt);
267 solve();
268 }
269
270 #ifndef ONLINE_JUDGE
271 printf("time = %d.\n", (int)clock());
272 #endif
273
274 return 0;
275 }
时间： 2024-10-25 00:40:34