P1172：二分，最大化最小值
1 #include <algorithm>
2 #include <bitset>
3 #include <cctype>
4 #include <complex>
5 #include <cstdio>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <string>
12 #include <vector>
13
14 using namespace std;
15
16 const int inf = 0x3f3f3f3f;
17
18 #define rep(id,from,to) for (int id = (from); id < (to); id++)
19 #define irep(id,from,to) for (int id = (from); id > (to); id--)
20 #define reset(arr,c) memset (arr, c, sizeof (arr) )
21 #define grt(type) greater<type>
22 #define Que(type) queue<type>
23 #define Pque(type) priority_queue<type>
24 #define gPque(type) priority_queue<type, vector<type>, greater<type> >
25
26 map<string, int> city;
27 int N, R;
28
29 struct Edge
30 {
31 int to, next, len;
32 void assign (int t, int n, int l)
33 {
34 to = t;
35 next = n;
36 len = l;
37 }
38 };
39
40 Edge elist[40000];
41 int head[205];
42 int ecnt;
43 int st, dest;
44
45 inline void add (int v1, int v2, int l)
46 {
47 elist[ecnt].assign (v2, head[v1], l);
48 head[v1] = (ecnt++);
49 elist[ecnt].assign (v1, head[v2], l);
50 head[v2] = (ecnt++);
51 }
52
53 bool input()
54 {
55 reset (head, -1);
56 ecnt = 0;
57 cin >> N >> R;
58 if (N == 0) return false;
59 string nm1, nm2;
60 int len;
61 int cnt = 0;
62 rep (i, 0, R)
63 {
64 cin >> nm1 >> nm2 >> len;
65 if (city.count (nm1) == 0) city[nm1] = (cnt++);
66 if (city.count (nm2) == 0) city[nm2] = (cnt++);
67 add (city[nm1], city[nm2], len);
68 }
69 cin >> nm1 >> nm2;
70 st = city[nm1];
71 dest = city[nm2];
72 return true;
73 }
74
75 bool vis[205];
76
77 bool dfs (int cur, int lim)
78 {
79 vis[cur] = true;
80 if (cur == dest) return true;
81 bool res = false;
82 for (int t, e = head[cur]; e != -1; e = elist[e].next)
83 {
84 t = elist[e].to;
85 if (!vis[t] && elist[e].len >= lim)
86 res |= dfs (t, lim);
87 }
88 return res;
89 }
90
91 int solve()
92 {
93 int lb = 1, rb = 10000;
94 while (lb < rb)
95 {
96 reset (vis, 0);
97 int m = (lb + rb + 1) >> 1;
98 if (!dfs (st, m) ) rb = m - 1;
99 else lb = m;
100 }
101 return lb;
102 }
103
104 int main()
105 {
106 int ks (0);
107 while (input() ) printf ("Scenario #%d\n%d tons\n\n", ++ks, solve() );
108 return 0;
109 }
TOJ P1172
P1290：模拟题
1 #include <algorithm>
2 #include <bitset>
3 #include <cctype>
4 #include <complex>
5 #include <cstdio>
6 #include <cstring>
7 #include <map>
8 #include <queue>
9 #include <set>
10 #include <vector>
11
12 using namespace std;
13
14 const int inf = 0x3f3f3f3f;
15
16 #define rep(id,from,to) for (int id = (from); id < (to); id++)
17 #define irep(id,from,to) for (int id = (from); id > (to); id--)
18 #define reset(arr,c) memset (arr, c, sizeof (arr) )
19 #define grt(type) greater<type>
20 #define Que(type) queue<type>
21 #define Pque(type) priority_queue<type>
22 #define gPque(type) priority_queue<type, vector<type>, greater<type> >
23
24
25 struct Card
26 {
27 char suit;
28 int val;
29 bool read()
30 {
31 char ch[2];
32 if (scanf ("%s", ch) == EOF) return false;
33 suit = ch[1];
34 if (ch[0] >= ‘2‘ && ch[0] <= ‘9‘) val = ch[0] - ‘0‘;
35 else switch (ch[0])
36 {
37 case ‘J‘:
38 val = 11;
39 break;
40 case ‘Q‘:
41 val = 12;
42 break;
43 case ‘K‘:
44 val = 13;
45 break;
46 case ‘A‘:
47 val = 14;
48 break;
49 }
50 return true;
51 }
52 };
53
54 int isStraight (int* cnt, bool isFlush)
55 {
56 bool ok = true;
57 int i;
58 rep (j, 2, 15) if (cnt[j])
59 {
60 i = j;
61 break;
62 }
63 rep (j, i + 1, i + 5) if (!cnt[j])
64 {
65 ok = false;
66 break;
67 }
68 return ok ? (isFlush ? 8 : 4) : 0;
69 }
70
71 int isFourOfAKind (int* cnt)
72 {
73 bool ok = false;
74 rep (i, 2, 15) if (cnt[i] == 4)
75 {
76 ok = true;
77 break;
78 }
79 return ok ? 7 : 0;
80 }
81
82 int isFullHouse (int* cnt)
83 {
84 bool f3 = false, f2 = false;
85 rep (i, 2, 15)
86 {
87 if (cnt[i] == 3) f3 = true;
88 else if (cnt[i] == 2) f2 = true;
89 }
90 return f3 && f2 ? 6 : 0;
91 }
92
93 int isThreeOfAKind (int* cnt)
94 {
95 rep (i, 2, 15) if (cnt[i] == 3) return 3;
96 return 0;
97 }
98
99 int isPair (int* cnt)
100 {
101 int res = 0;
102 rep (i, 2, 15) if (cnt[i] == 2) ++res;
103 return res;
104 }
105
106 int cmpStraight (int* cntA, int* cntB)
107 {
108 int ma, mb;
109 irep (i, 14, 1) if (cntA[i])
110 {
111 ma = i;
112 break;
113 }
114 irep (i, 14, 1) if (cntB[i])
115 {
116 mb = i;
117 break;
118 }
119 if (ma > mb) return 1;
120 else if (ma < mb) return -1;
121 else return 0;
122 }
123
124 int cmpFour (int* cntA, int* cntB)
125 {
126 int ma, mb;
127 rep (i, 2, 15) if (cntA[i] == 4)
128 {
129 ma = i;
130 break;
131 }
132 rep (i, 2, 15) if (cntB[i] == 4)
133 {
134 mb = i;
135 break;
136 }
137 if (ma > mb) return 1;
138 else if (mb > ma) return -1;
139 else return 0;
140 }
141
142 int cmpFullHouse (int* cntA, int* cntB)
143 {
144 int ma, mb;
145 rep (i, 2, 15) if (cntA[i] == 3)
146 {
147 ma = i;
148 break;
149 }
150 rep (i, 2, 15) if (cntB[i] == 3)
151 {
152 mb = i;
153 break;
154 }
155 if (ma > mb) return 1;
156 else if (mb > ma) return -1;
157 else return 0;
158 }
159
160 int cmpPair (int* cntA, int* cntB, int pr)
161 {
162 int pa = 0, ha = 0, pb = 0, hb = 0;
163 int ma[5] = {0}, mb[5] = {0};
164 irep (i, 14, 1)
165 {
166 if (cntA[i] == 2) ma[pa++] = i;
167 else if (cntA[i] == 1) ma[pr + (ha++)] = i;
168 }
169 irep (i, 14, 1)
170 {
171 if (cntB[i] == 2) mb[pb++] = i;
172 else if (cntB[i] == 1) mb[pr + (hb++)] = i;
173 }
174 rep (i, 0, 5)
175 {
176 if (ma[i] > mb[i]) return 1;
177 else if (mb[i] > ma[i]) return -1;
178 }
179 return 0;
180 }
181
182 int cmp (Card* A, Card* B)
183 {
184 int cntA[15] = {0};
185 int cntB[15] = {0};
186 rep (i, 0, 5) ++cntA[A[i].val];
187 rep (i, 0, 5) ++cntB[B[i].val];
188 int typeA = 0, typeB = 0;
189 bool isFlushA = true, isFlushB = true;
190 rep (i, 1, 5) if (A[i].suit != A[0].suit)
191 {
192 isFlushA = false;
193 break;
194 }
195 rep (i, 1, 5) if (B[i].suit != B[0].suit)
196 {
197 isFlushB = false;
198 break;
199 }
200 typeA = isFlushA ? 5 : 0;
201 typeA = max (typeA, isStraight (cntA, isFlushA) );
202 typeA = max (typeA, isFourOfAKind (cntA) );
203 typeA = max (typeA, isFullHouse (cntA) );
204 typeA = max (typeA, isThreeOfAKind (cntA) );
205 typeA = max (typeA, isPair (cntA) );
206 typeB = isFlushA ? 5 : 0;
207 typeB = max (typeB, isStraight (cntB, isFlushB) );
208 typeB = max (typeB, isFourOfAKind (cntB) );
209 typeB = max (typeB, isFullHouse (cntB) );
210 typeB = max (typeB, isThreeOfAKind (cntB) );
211 typeB = max (typeB, isPair (cntB) );
212 if (typeA > typeB) return 1;
213 else if (typeA < typeB) return -1;
214 if (typeA == 4 || typeA == 8) return cmpStraight (cntA, cntB);
215 else if (typeA == 7) return cmpFour (cntA, cntB);
216 else if (typeA == 6 || typeA == 3) return cmpFullHouse (cntA, cntB);
217 else return cmpPair (cntA, cntB, (typeA == 5) ? 0 : typeA);
218 return 0;
219 }
220
221 int main()
222 {
223 Card A[5], B[5];
224 while (A[0].read() )
225 {
226 rep (i, 1, 5) A[i].read();
227 rep (i, 0, 5) B[i].read();
228 int ans = cmp (A, B);
229 if (ans == 1) printf ("Black wins.\n");
230 else if (ans == -1) printf ("White wins.\n");
231 else printf ("Tie.\n");
232 }
233 return 0;
234 }
TOJ P1290
P1557：H2O，暴力枚举即可
1 #include <algorithm>
2 #include <bitset>
3 #include <cctype>
4 #include <complex>
5 #include <cstdio>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <string>
12 #include <vector>
13
14 using namespace std;
15
16 const int inf = 0x3f3f3f3f;
17
18 #define rep(id,from,to) for (int id = (from); id < (to); id++)
19 #define irep(id,from,to) for (int id = (from); id > (to); id--)
20 #define reset(arr,c) memset (arr, c, sizeof (arr) )
21 #define grt(type) greater<type>
22 #define Que(type) queue<type>
23 #define Pque(type) priority_queue<type>
24 #define gPque(type) priority_queue<type, vector<type>, greater<type> >
25
26 int pr[16], pg[16], pb[16];
27
28 inline int dist (int r, int g, int b, int id)
29 {
30 return (r - pr[id]) * (r - pr[id]) + (g - pg[id]) * (g - pg[id]) + (b - pb[id]) * (b - pb[id]);
31 }
32
33 int main()
34 {
35 rep (i, 0, 16) scanf ("%d%d%d", pr + i, pg + i, pb + i);
36 int r, g, b;
37 while (1)
38 {
39 scanf ("%d%d%d", &r, &g, &b);
40 if (r == -1) break;
41 int md = inf, id = inf;
42 rep (i, 0, 16)
43 {
44 int d = dist (r, g, b, i);
45 if (d < md)
46 {
47 md = d;
48 id = i;
49 }
50 }
51 printf ("(%d,%d,%d) maps to (%d,%d,%d)\n", r, g, b, pr[id], pg[id], pb[id]);
52 }
53 return 0;
54 }
TOJ P1557
时间： 2024-10-18 17:35:16