D1T1> 神奇的幻方
模拟即可。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #define rep(i, a, b) for (int i = a; i <= b; i++)
5 #define drep(i, a, b) for (int i = a; i >= b; i--)
6 #define REP(i, a, b) for (int i = a; i < b; i++)
7 #define pb push_back
8 #define mp make_pair
9 #define clr(x) memset(x, 0, sizeof(x))
10 #define xx first
11 #define yy second
12
13 using namespace std;
14
15 typedef long long i64;
16 typedef pair<int, int> pii;
17 const int inf = ~0U >> 1;
18 const i64 INF = ~0ULL >> 1;
19 //******************************
20
21 int map[55][55];
22 int main() {
23 int n;
24 scanf("%d", &n);
25 pii last = mp(1, n / 2 + 1);
26 map[1][n / 2 + 1] = 1;
27 rep(i, 2, n * n) {
28 if (last.xx == 1 && last.yy != n) {
29 map[n][last.yy + 1] = i;
30 last = mp(n, last.yy + 1);
31 }
32 else if (last.yy == n && last.xx != 1) {
33 map[last.xx - 1][1] = i;
34 last = mp(last.xx - 1, 1);
35 }
36 else if (last.xx == 1 && last.yy == n) {
37 map[last.xx + 1][last.yy] = i;
38 last = mp(last.xx + 1, last.yy);
39 }
40 else if (last.xx != 1 && last.yy != n && !map[last.xx - 1][last.yy + 1]) {
41 map[last.xx - 1][last.yy + 1] = i;
42 last = mp(last.xx - 1, last.yy + 1);
43 }
44 else {
45 map[last.xx + 1][last.yy] = i;
46 last = mp(last.xx + 1, last.yy);
47 }
48 }
49 rep(i, 1, n) {
50 rep(j, 1, n) {
51 if (j != 1) printf(" ");
52 printf("%d", map[i][j]);
53 }
54 puts("");
55 }
56 return 0;
57 }
D1T2> 信息传递
一个点只有一个出度 => 环不会交叉
1)dfs找环 2)tarjan找scc
考试的时候就直接写了tarjan...
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #define rep(i, a, b) for (int i = a; i <= b; i++)
5 #define drep(i, a, b) for (int i = a; i >= b; i--)
6 #define REP(i, a, b) for (int i = a; i < b; i++)
7 #define pb push_back
8 #define mp make_pair
9 #define clr(x) memset(x, 0, sizeof(x))
10 #define xx first
11 #define yy second
12
13 using namespace std;
14
15 typedef long long i64;
16 typedef pair<int, int> pii;
17 const int inf = ~0U >> 1;
18 const i64 INF = ~0ULL >> 1;
19 //******************************
20
21 const int maxn = 200005;
22
23 struct Ed {
24 int u, v, nx; Ed() {}
25 Ed(int _u, int _v, int _nx)
26 :u(_u), v(_v), nx(_nx) {}
27 } a[maxn];
28 int fst[maxn], cnt;
29 void addedge(int u, int v) {
30 a[++cnt] = Ed(u, v, fst[u]);
31 fst[u] = cnt;
32 }
33
34 int dfn[maxn], low[maxn], Index, vis[maxn];
35 int stack[maxn], top;
36 int scc_cnt;
37 int belong[maxn], size[maxn];
38 void tarjan(int u) {
39 dfn[u] = low[u] = ++Index;
40 vis[u] = 1;
41 stack[++top] = u;
42 for (int i = fst[u]; i; i = a[i].nx) {
43 int v = a[i].v;
44 if (!dfn[v]) {
45 tarjan(v);
46 low[u] = min(low[u], low[v]);
47 }
48 else if (vis[v]) {
49 low[u] = min(low[u], dfn[v]);
50 }
51 }
52 if (dfn[u] == low[u]) {
53 scc_cnt++;
54 while (stack[top] != u) {
55 belong[stack[top]] = scc_cnt;
56 vis[stack[top]] = 0;
57 size[scc_cnt]++;
58 top--;
59 }
60 belong[stack[top]] = scc_cnt;
61 vis[stack[top]] = 0;
62 size[scc_cnt]++;
63 top--;
64 }
65 }
66
67 int main() {
68 int n;
69 scanf("%d", &n);
70 rep(i, 1, n) {
71 int id;
72 scanf("%d", &id);
73 addedge(i, id);
74 }
75
76 rep(i, 1, n) {
77 if (!dfn[i]) tarjan(i);
78 }
79
80 int ans = 0x3f3f3f3f;
81 rep(i, 1, scc_cnt) if (size[i] != 1) ans = min(ans, size[i]);
82
83 printf("%d\n", ans);
84 return 0;
85 }
D1T3> 斗地主
搜索题，以前在contest hunter上见过，没有订正...
因为1 2 3 4 5不能顺, 1 1 1 2 2 2不能连出，可以在读入时将牌的编号重新分配，只需要将 A 调整到 K 后，2 和 A 断开即可。
考试的时候想把所有可行的方案预处理出来，压位压在一起，然后状压dp，这样是不可行的，因为方案数多了后时间复杂度会爆炸。
以下代码可以过NOIP官方数据，但是过不了UOJ的加强版题目，会WA，这篇代码的思路主要是在每次判断时，人为地把4张和3张出掉，强剪枝。
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (register int i = a; i <= b; i++)
3 #define drep(i, a, b) for (register int i = a; i >= b; i--)
4 #define REP(i, a, b) for (int i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define clr(x) memset(x, 0, sizeof(x))
8 #define xx first
9 #define yy second
10
11 using namespace std;
12
13 typedef long long i64;
14 typedef pair<int, int> pii;
15 const int inf = ~0U >> 1;
16 const i64 INF = ~0ULL >> 1;
17 //********************************
18
19 const int maxn = 20;
20
21 inline int read() {
22 int l = 1, s = 0;
23 char ch = getchar();
24 while (ch < ‘0‘ || ch > ‘9‘) { if (ch == ‘-‘) l = -1; ch = getchar(); }
25 while (ch >= ‘0‘ && ch <= ‘9‘) { s = (s << 3) + (s << 1) + ch - ‘0‘; ch = getchar(); }
26 return l * s;
27 }
28
29 int poker[maxn];
30
31 int ans = inf;
32
33 void dfs(int);
34
35 void group(int base, int len, int dep) {
36 rep(st, 1, 13 - len) {
37 if (poker[st] < base) continue;
38 int end;
39 for (end = st; end <= 12; end++) if (poker[end] < base) break;
40 if (--end - st + 1 < len) continue;
41 drep(k, end, st + len - 1) {
42 rep(i, st, k) poker[i] -= base;
43 poker[0] -= (k - st + 1) * base;
44 dfs(dep + 1);
45 rep(i, st, k) poker[i] += base;
46 poker[0] += (k - st + 1) * base;
47 }
48 }
49 }
50
51 int n, tong[maxn];
52
53 void dfs(int dep) {
54 if (dep > ans) return;
55 if (poker[0] == 0) return;
56 rep(i, 1, 16) tong[poker[i]]++;
57 int sum(0);
58 while (tong[4] > 0) {
59 sum++;
60 tong[4]--;
61 if (tong[1] >= 2) tong[1] -= 2;
62 else if (tong[2] >= 2) tong[2] -= 2;
63 }
64 while (tong[3] > 0) {
65 sum++;
66 tong[3]--;
67 if (tong[1] >= 1) tong[1] -= 1;
68 else if (tong[2] >= 1) tong[2] -= 1;
69 }
70 if (tong[1] >= 2 && poker[15] >= 1 && poker[16] >= 1) sum--;
71 if (sum + tong[1] + tong[2] + dep < ans) ans = sum + tong[1] + tong[2] + dep;
72 tong[1] = tong[2] = 0;
73
74 group(3, 2, dep);
75 group(2, 3, dep);
76 group(1, 5, dep);
77 }
78
79 int main() {
80 int T;
81 T = read(), n = read();
82 while (T--) {
83 memset(poker, 0, sizeof(poker));
84 rep(i, 1, n) {
85 int rb, id;
86 id = read(); rb = read();
87 if (id >= 3 && id <= 13) poker[id - 2]++;
88 else if (id == 1) poker[12]++;
89 else if (id == 2) poker[14]++;
90 else if (id == 0 && poker[15] == 0) poker[15]++;
91 else poker[16]++;
92 }
93 poker[0] = n;
94 ans = inf;
95 dfs(0);
96 printf("%d\n", ans);
97 }
98 return 0;
99 }
这一篇代码依然可以过NOIP官方数据，但是在UOJ加强版上在第9个extra test上t了，不知道卡时是否可以。
这个思路考虑了 AAAA2222 在一次出完的情况， 还考虑了在4带2时可以破坏3个一样的牌的情况，分别写出来就好了。
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (int i = a; i <= b; i++)
3 #define drep(i, a, b) for (int i = a; i >= b; i--)
4 #define REP(i, a, b) for (int i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define clr(x) memset(x, 0, sizeof(x))
8 #define xx first
9 #define yy second
10
11 using namespace std;
12
13 typedef long long i64;
14 typedef pair<int, int> pii;
15 const int inf = ~0U >> 1;
16 const i64 INF = ~0ULL >> 1;
17 //********************************
18
19 const int maxn = 20;
20
21 int poker[maxn];
22
23 int ans = inf;
24
25 void dfs(int dep, int order) {
26 if (dep > ans) return;
27 if (dep + poker[0] < ans) ans = dep + poker[0];
28 if (poker[0] == 0) return;
29
30 //AAABBBCCC
31 if (order <= 0)
32 rep(st, 1, 12) {
33 if (poker[st] < 3) continue;
34 int end;
35 for (end = st; end <= 12; end++) if (poker[end] < 3) break;
36 end--; if (end - st < 1) break;
37 rep(k, st + 1, end) {
38 rep(i, st, k) poker[i] -= 3;
39 poker[0] -= (k - st + 1) * 3;
40 dfs(dep + 1, 0);
41 rep(i, st, k) poker[i] += 3;
42 poker[0] += (k - st + 1) * 3;
43 }
44 }
45
46 //AABBCC
47 if (order <= 1)
48 rep(st, 1, 12) {
49 if (poker[st] < 2) continue;
50 int end;
51 for (end = st; end <= 12; end++) if (poker[end] < 2) break;
52 end--; if (end - st + 1 < 3) continue;
53 rep(k, st + 2, end) {
54 rep(i, st, k) poker[i] -= 2;
55 poker[0] -= (k - st + 1) * 2;
56 dfs(dep + 1, 1);
57 rep(i, st, k) poker[i] += 2;
58 poker[0] += (k - st + 1) * 2;
59 }
60 }
61
62 //ABCDE
63 if (order <= 2)
64 rep(st, 1, 12) {
65 if (poker[st] < 1) continue;
66 int end;
67 for (end = st; end <= 12; end++) if (poker[end] < 1) break;
68 end--; if (end - st + 1 < 5) continue;
69 rep(k, st + 4, end) {
70 rep(i, st, k) poker[i] -= 1;
71 poker[0] -= k - st + 1;
72 dfs(dep + 1, 2);
73 rep(i, st, k) poker[i] += 1;
74 poker[0] += k - st + 1;
75 }
76 }
77
78 //AAAABC
79 if (order <= 3)
80 rep(st, 1, 14) {
81 if (poker[st] < 4) continue;
82 rep(i, 1, 16) {
83 if (poker[i] < 1 || i == st) continue;
84 poker[i] -= 1;
85 rep(j, i, 16) {
86 if (poker[j] < 1 || j == st) continue;
87 poker[st] -= 4; poker[j] -= 1;
88 poker[0] -= 6;
89 dfs(dep + 1, 3);
90 poker[st] += 4; poker[j] += 1;
91 poker[0] += 6;
92 }
93 poker[i] += 1;
94 }
95 }
96
97 //AAAABBCC
98 if (order <= 4)
99 rep(st, 1, 14) {
100 if (poker[st] < 4) continue;
101 rep(i, 1, 14) {
102 if (poker[i] < 2 || i == st) continue;
103 poker[i] -= 2;
104 rep(j, i, 14) {
105 if (poker[j] < 2 || j == st) continue;
106 poker[st] -= 4, poker[j] -= 2;
107 poker[0] -= 8;
108 dfs(dep + 1, 4);
109 poker[st] += 4, poker[j] += 2;
110 poker[0] += 8;
111 }
112 poker[i] += 2;
113 }
114 }
115
116 //AAABB
117 if (order <= 5)
118 rep(st, 1, 14) {
119 if (poker[st] < 3) continue;
120 rep(i, 1, 14) {
121 if (poker[i] < 2 || i == st) continue;
122 poker[st] -= 3, poker[i] -= 2;
123 poker[0] -= 5;
124 dfs(dep + 1, 5);
125 poker[st] += 3, poker[i] += 2;
126 poker[0] += 5;
127 }
128 }
129
130 //AAAB
131 if (order <= 6)
132 rep(st, 1, 14) {
133 if (poker[st] < 3) continue;
134 rep(i, 1, 16) {
135 if (poker[i] < 1 || i == st) continue;
136 poker[st] -= 3, poker[i] -= 1;
137 poker[0] -= 4;
138 dfs(dep + 1, 6);
139 poker[st] += 3, poker[i] += 1;
140 poker[0] += 4;
141 }
142 }
143
144 //AAAA
145 if (order <= 7)
146 rep(st, 1, 14) {
147 if (poker[st] < 4) continue;
148 poker[st] -= 4, poker[0] -= 4;
149 dfs(dep + 1, 7);
150 poker[st] += 4, poker[0] += 4;
151 }
152
153 //AAA
154 if (order <= 8)
155 rep(st, 1, 14) {
156 if (poker[st] < 3) continue;
157 poker[st] -= 3, poker[0] -= 3;
158 dfs(dep + 1, 8);
159 poker[st] += 3, poker[0] += 3;
160 }
161
162 //AA
163 if (order <= 9)
164 rep(st, 1, 14) {
165 if (poker[st] < 2) continue;
166 poker[st] -= 2, poker[0] -= 2;
167 dfs(dep + 1, 9);
168 poker[st] += 2, poker[0] += 2;
169 }
170
171 //Joker
172 if (order <= 10)
173 if (poker[15] && poker[16]) {
174 poker[15] = poker[16] = 0;
175 poker[0] -= 2;
176 dfs(dep + 1, 10);
177 poker[15] = poker[16] = 1;
178 poker[0] += 2;
179 }
180 }
181
182 int main() {
183 int T, n;
184 scanf("%d%d", &T, &n);
185 while (T--) {
186 memset(poker, 0, sizeof(poker));
187 rep(i, 1, n) {
188 int rb, id;
189 scanf("%d%d", &id, &rb);
190 if (id >= 3 && id <= 13) poker[id - 2]++;
191 else if (id == 1) poker[12]++;
192 else if (id == 2) poker[14]++;
193 else if (id == 0 && poker[15] == 0) poker[15]++;
194 else poker[16]++;
195 }
196 poker[0] = n;
197 ans = inf;
198 dfs(0, 0);
199 printf("%d\n", ans);
200 }
201 return 0;
202 }
D2T1> 跳石头
二分+贪心。
noi.openjudge.cn的原题，首先二分答案，在二分答案的情况下，扫一遍n个石头，如果这个石头与前一个石头的距离小于二分的答案，那么就把前一个石头取消。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #define rep(i, a, b) for (int i = a; i <= b; i++)
5 #define drep(i, a, b) for (int i = a; i >= b; i--)
6 #define REP(i, a, b) for (int i = a; i < b; i++)
7 #define pb push_back
8 #define mp make_pair
9 #define clr(x) memset(x, 0, sizeof(x))
10 #define xx first
11 #define yy second
12
13 using namespace std;
14
15 typedef long long i64;
16 typedef pair<int, int> pii;
17 const int inf = ~0U >> 1;
18 const i64 INF = ~0ULL >> 1;
19 //******************************
20
21 int map[55][55];
22 int main() {
23 int n;
24 scanf("%d", &n);
25 pii last = mp(1, n / 2 + 1);
26 map[1][n / 2 + 1] = 1;
27 rep(i, 2, n * n) {
28 if (last.xx == 1 && last.yy != n) {
29 map[n][last.yy + 1] = i;
30 last = mp(n, last.yy + 1);
31 }
32 else if (last.yy == n && last.xx != 1) {
33 map[last.xx - 1][1] = i;
34 last = mp(last.xx - 1, 1);
35 }
36 else if (last.xx == 1 && last.yy == n) {
37 map[last.xx + 1][last.yy] = i;
38 last = mp(last.xx + 1, last.yy);
39 }
40 else if (last.xx != 1 && last.yy != n && !map[last.xx - 1][last.yy + 1]) {
41 map[last.xx - 1][last.yy + 1] = i;
42 last = mp(last.xx - 1, last.yy + 1);
43 }
44 else {
45 map[last.xx + 1][last.yy] = i;
46 last = mp(last.xx + 1, last.yy);
47 }
48 }
49 rep(i, 1, n) {
50 rep(j, 1, n) {
51 if (j != 1) printf(" ");
52 printf("%d", map[i][j]);
53 }
54 puts("");
55 }
56 return 0;
57 }
D2T2> 子串
dp。
用f[i][j][k][2]表示第一个字符串中的前i位，取出k段，按顺序组合，拼出第二个字符串前j位的方案数，最后一位若为1则表示第i位被考虑在内(被选取)，为0则为不被选取。
那么当S[i] == T[j]时(S 为 第一个字符串，T 为 第二个字符串)
f[i][j][k][1] = f[i - 1][j - 1][k][1] + f[i - 1][j - 1][k - 1][0] + f[i - 1][j - 1][k - 1][1];
f[i][j][k][0] = f[i - 1][j][k][0];
否则 f[i][j][k][1] = 0, f[i][j][k][0] = f[i - 1][j][k][0] + f[i - 1][j][k][1];
解释:当前字符能否接下去(假若S[i] ==T[j])实则取决于第i - 1个字符是否选取。
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (int i = a; i <= b; i++)
3 #define drep(i, a, b) for (int i = a; i >= b; i--)
4 #define REP(i, a, b) for (int i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define clr(x) memset(x, 0, sizeof(x))
8 #define xx first
9 #define yy second
10
11 using namespace std;
12
13 typedef long long i64;
14 typedef pair<int, int> pii;
15 const int inf = ~0U >> 1;
16 const i64 INF = ~0ULL >> 1;
17 //*******************************
18
19 const int maxn = 1005, maxm = 205;
20 const int mod = 1000000007;
21
22 int read() {
23 int s = 0, l = 1;
24 char ch = getchar();
25 while (ch < ‘0‘ || ch > ‘9‘) { if (ch == ‘-‘) l = -1; ch = getchar(); }
26 while (ch >= ‘0‘ && ch <= ‘9‘) { s = (s << 3) + (s << 1) + ch - ‘0‘; ch = getchar(); }
27 return l * s;
28 }
29
30 char str1[maxn], str2[maxm];
31 i64 f[2][maxm][maxm][2];
32
33 int main() {
34 int n, m, k;
35 n = read(), m = read(), k = read();
36 scanf("%s%s", str1 + 1, str2 + 1);
37 f[0][0][0][0] = f[1][0][0][0] = 1;
38 int flag = 0;
39 rep(i, 1, n) {
40 flag ^= 1;
41 int up = min(i, m);
42 rep(j, 1, up) {
43 if (str1[i] == str2[j]) rep(kk, 1, j) {
44 f[flag][j][kk][0] = (f[1 ^ flag][j][kk][0] + f[1 ^ flag][j][kk][1]) % mod;
45 f[flag][j][kk][1] = (f[1 ^ flag][j - 1][kk][1] + f[1 ^ flag][j - 1][kk - 1][0] + f[1 ^ flag][j - 1][kk - 1][1]) % mod;
46 }
47 else rep(kk, 1, j) {
48 f[flag][j][kk][0] = (f[1 ^ flag][j][kk][0] + f[1 ^ flag][j][kk][1]) % mod;
49 f[flag][j][kk][1] = 0;
50 }
51 }
52 }
53 printf("%lld\n", (f[flag][m][k][0] + f[flag][m][k][1]) % mod);
54 return 0;
55 }
D2T3> 运输计划
两天内改完......
总结:考得挺扯蛋的，335，非常不满意:(
暴力分没有拿全，对算法的掌握不够深刻，最恶心的是做过的题也不会。
这TM混蛋的人生又干掉了我的希望，就这样吧，也没什么好说的...
时间： 2024-10-26 23:08:30