zoukankan      html  css  js  c++  java

  • 第一周 1.17-1.23

    1.17

    CF 614 E Necklace

    类似于函数图像有两条对称轴 必然是周期的。

    如果循环次数是奇数 每一循环节都要是回文 把奇数的放中间 偶数放两边 （所以奇数的不止1个就无解）。

    循环次数偶 随便放一个循环节 相邻的对称一下即可。

     1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 const int maxn = 1e5 + 10;
 5 int num[26];
 6 char s[maxn];
 7 
 8 int gcd(int a, int b)
 9 {
10     return a % b ? gcd(b, a % b) : b;
11 }
12 
13 int main(void)
14 {
15     int n, cnt = 0;
16     scanf("%d", &n);
17     for(int i = 0; i < n; i++) scanf("%d", num + i);
18     int tmp = num[0];
19     for(int i = 1; i < n; i++) tmp = gcd(tmp, num[i]);
20     for(int i = 0; i < n; i++) num[i] /= tmp;
21     if(tmp & 1)
22     {
23         int pos1 = -1;
24         for(int i = 0; i < n; i++)
25         {
26             if(num[i] & 1)
27             {
28                 if(pos1 == -1) pos1 = i;
29                 else
30                 {
31                     puts("0");
32                     for(int k = 0; k < n; k++)
33                         for(int j = 0; j < tmp * num[k]; j++)
34                             putchar('a' + k);
35                     puts("");
36                     return 0;
37                 }
38             }
39         }
40         for(int i = 0; i < n; i++)
41         {
42             if(num[i] & 1) num[i]--;
43             for(int j = 0; j < num[i] / 2; j++)
44                 s[cnt++] = 'a' + i;
45         }
46         int l = cnt;
47         if(pos1 != -1) s[cnt++] = 'a' + pos1;
48         for(int j = 1; j <= l; j++) s[cnt++] = s[l-j];
49         s[cnt] = 0;
50         printf("%d
", tmp);
51         for(int j = 0; j < tmp; j++) printf("%s", s);
52         puts("");
53     }
54     else
55     {
56         for(int i = 0; i < n; i++)
57             for(int j = 0; j < num[i]; j++)
58                 s[cnt++] = 'a' + i;
59         for(int i = cnt; i; i--)
60             s[cnt++] = s[i-1];
61         s[cnt] = 0;
62         printf("%d
", tmp);
63         for(int j = 0; j < tmp / 2; j++) printf("%s", s);
64         puts("");
65     }
66     return 0;
67 }
    Aguin

    原来看后缀数组是上个月的事情了？

    POJ 2217 Secretary

    最长公共子串。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 22222;
 7 char s[maxn];
 8 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn];
 9 bool cmp(int i, int j)
10 {
11     if(r[i] != r[j]) return r[i] < r[j];
12     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
13 }
14 void get_SA()
15 {
16     for(int i = 0; i <= n; i++)
17     {
18         SA[i] = i;
19         r[i] = i < n ? s[i] : -1;
20     }
21     for(k = 1; k <= n; k <<= 1)
22     {
23         sort(SA, SA + n + 1, cmp);
24         tmp[SA[0]] = 0;
25         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
26         memcpy(r, tmp, sizeof(r));
27     }
28     return;
29 }
30 void get_height()
31 {
32     for(int i = 0; i <= n; i++) r[SA[i]] = i;
33     int k = 0;
34     for(int i = 0; i < n; i++)
35     {
36         if(k) k--;
37         int j = SA[r[i]-1];
38         while(s[i+k] == s[j+k]) k++;
39         h[r[i]] = k;
40     }
41     return;
42 }
43 
44 int main(void)
45 {
46     int N;
47     scanf("%d", &N);
48     getchar();
49     while(N--)
50     {
51         gets(s);
52         int m = strlen(s);
53         s[m] = '~';
54         gets(s + m + 1);
55         n = strlen(s);
56         get_SA();
57         get_height();
58         int ans = 0;
59         for(int i = 1; i <= n; i++)
60             if( ( SA[i] <= m ) != ( SA[i-1] <= m ) && h[i] > ans)
61                 ans = h[i];
62         printf("Nejdelsi spolecny retezec ma delku %d.
", ans);
63     }
64     return 0;
65 }
    Aguin

    UVA 11107 Life Forms

    讨厌UVA。

      1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 const int maxn = 111111;
  7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn];
  8 int range[maxn], flag[111];
  9 char s[maxn];
 10 
 11 bool cmp(int i, int j)
 12 {
 13     if(r[i] != r[j]) return r[i] < r[j];
 14     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
 15 }
 16 
 17 void get_SA()
 18 {
 19     for(int i = 0; i <= n; i++)
 20     {
 21         SA[i] = i;
 22         r[i] = i < n ? s[i] : -1;
 23     }
 24     for(k = 1; k <= n; k <<= 1)
 25     {
 26         sort(SA, SA + n + 1, cmp);
 27         tmp[SA[0]] = 0;
 28         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
 29         memcpy(r, tmp, sizeof(r));
 30     }
 31     return;
 32 }
 33 
 34 void get_height()
 35 {
 36     for(int i = 0; i <= n; i++) r[SA[i]] = i;
 37     int k = 0;
 38     for(int i = 0; i < n; i++)
 39     {
 40         if(k) k--;
 41         int j = SA[r[i]-1];
 42         while(s[i+k] == s[j+k]) k++;
 43         h[r[i]] = k;
 44     }
 45     return;
 46 }
 47 
 48 int main(void)
 49 {
 50     int N, fir = 1;
 51     while(~scanf("%d", &N) && N)
 52     {
 53         if(!fir) puts("");
 54         fir = 0;
 55         if(N == 1)
 56         {
 57             scanf("%s", s);
 58             printf("%s

", s);
 59             continue;
 60         }
 61         int pos = 0;
 62         for(int i = 1; i <= N; i++)
 63         {
 64             scanf("%s", s + pos);
 65             for(int j = pos; s[j]; j++) range[j] = i;
 66             pos = strlen(s);
 67             s[pos++] = 130 + i;
 68         }
 69         s[pos] = 0;
 70         n = strlen(s);
 71         get_SA();
 72         get_height();
 73         int l = 0, r = n, mid;
 74         while(l < r)
 75         {
 76             int ok = 0;
 77             mid = (l + r + 1) / 2;
 78             for(int i = 1; i <= n; i++)
 79             {
 80                 if(h[i] < mid) continue;
 81                 int cnt = 0;
 82                 memset(flag, 0, sizeof(flag));
 83                 for(int j = i; j <= n; j++)
 84                 {
 85                     if(h[j] < mid) break;
 86                     int a = range[SA[j]], b = range[SA[j-1]];
 87                     if(a == b) continue;
 88                     if(!flag[a]) cnt++;
 89                     if(!flag[b]) cnt++;
 90                     flag[a] = flag[b] = 1;
 91                 }
 92                 if(cnt > N / 2) {ok = 1;break;}
 93             }
 94             if(ok) l = mid;
 95             else r = mid - 1;
 96         }
 97         if(l) for(int i = 1; i <= n; i++)
 98         {
 99             if(h[i] < l) continue;
100             int cnt = 0;
101             memset(flag, 0, sizeof(flag));
102             for(int j = i; j <= n; j++)
103             {
104                 if(h[j] < l) break;
105                 int a = range[SA[j]], b = range[SA[j-1]];
106                 if(!a || !b || a == b) continue;
107                 if(!flag[a]) cnt++;
108                 if(!flag[b]) cnt++;
109                 flag[a] = flag[b] = 1;
110                 i = j;
111             }
112             if(cnt > N / 2)
113             {
114                 for(int j = 0; j < l; j++) putchar(s[SA[i]+j]);
115                 puts("");
116             }
117         }
118         else puts("?");
119     }
120     return 0;
121 }
    Aguin

    POJ 1743 Musical Theme

    做差搞SA。相邻要大于mid而不是大于等于mid才不重叠。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 111111;
 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn];
 8 int s[maxn];
 9 
10 bool cmp(int i, int j)
11 {
12     if(r[i] != r[j]) return r[i] < r[j];
13     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
14 }
15 
16 void get_SA()
17 {
18     for(int i = 0; i <= n; i++)
19     {
20         SA[i] = i;
21         r[i] = i < n ? s[i] : -1;
22     }
23     for(k = 1; k <= n; k <<= 1)
24     {
25         sort(SA, SA + n + 1, cmp);
26         tmp[SA[0]] = 0;
27         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
28         memcpy(r, tmp, sizeof(r));
29     }
30     return;
31 }
32 
33 void get_height()
34 {
35     for(int i = 0; i <= n; i++) r[SA[i]] = i;
36     int k = 0;
37     for(int i = 0; i < n; i++)
38     {
39         if(k) k--;
40         int j = SA[r[i]-1];
41         while(s[i+k] == s[j+k]) k++;
42         h[r[i]] = k;
43     }
44     return;
45 }
46 
47 int main(void)
48 {
49     int N;
50     while(~scanf("%d", &N) && N)
51     {
52         for(int i = 0; i < N; i++) scanf("%d", s + i);
53         for(int i = 1; i < N; i++) s[i-1] -= s[i];
54         n = N - 1;
55         get_SA();
56         get_height();
57         int l = 0, r = n - 1, mid;
58         while(l < r)
59         {
60             int ok = 0;
61             mid = (l + r + 1) / 2;
62             for(int i = 1; i < n; i++) if(h[i] >= mid)
63             {
64                 int Max = SA[i-1], Min = SA[i-1];
65                 for(int j = i; j < n; j++)
66                 {
67                     if(h[j] < mid) break;
68                     Max = max(Max, SA[j]);
69                     Min = min(Min, SA[j]);
70                     i = j;
71                 }
72                 if(Max - Min > mid) {ok = 1;break;}
73             }
74             if(ok) l = mid;
75             else r = mid - 1;
76         }
77         if(l > 3) printf("%d
", l + 1);
78         else puts("0");
79     }
80     return 0;
81 }
    Aguin

    POJ 3261 Milk Patterns

    上面改改就好啦。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 111111;
 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn];
 8 int s[maxn];
 9 
10 bool cmp(int i, int j)
11 {
12     if(r[i] != r[j]) return r[i] < r[j];
13     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
14 }
15 
16 void get_SA()
17 {
18     for(int i = 0; i <= n; i++)
19     {
20         SA[i] = i;
21         r[i] = i < n ? s[i] : -1;
22     }
23     for(k = 1; k <= n; k <<= 1)
24     {
25         sort(SA, SA + n + 1, cmp);
26         tmp[SA[0]] = 0;
27         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
28         memcpy(r, tmp, sizeof(r));
29     }
30     return;
31 }
32 
33 void get_height()
34 {
35     for(int i = 0; i <= n; i++) r[SA[i]] = i;
36     int k = 0;
37     for(int i = 0; i < n; i++)
38     {
39         if(k) k--;
40         int j = SA[r[i]-1];
41         while(s[i+k] == s[j+k]) k++;
42         h[r[i]] = k;
43     }
44     return;
45 }
46 
47 int main(void)
48 {
49     int K;
50     while(~scanf("%d%d", &n, &K))
51     {
52         for(int i = 0; i < n; i++) scanf("%d", s + i);
53         get_SA();
54         get_height();
55         int l = 0, r = n, mid;
56         while(l < r)
57         {
58             int ok = 0;
59             mid = (l + r + 1) / 2;
60             for(int i = 1; i <= n; i++) if(h[i] >= mid)
61             {
62                 int cnt = 1;
63                 for(int j = i; j <= n; j++)
64                 {
65                     if(h[j] < mid) break;
66                     cnt++;
67                     i = j;
68                 }
69                 if(cnt >= K) {ok = 1;break;}
70             }
71             if(ok) l = mid;
72             else r = mid - 1;
73         }
74         printf("%d
", l);
75     }
76     return 0;
77 }
    Aguin

    SPOJ DISUBSTR Distinct Substrings

    每个后缀对答案的贡献是n - sa[i] - height[i]。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 111111;
 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn];
 8 char s[maxn];
 9 
10 bool cmp(int i, int j)
11 {
12     if(r[i] != r[j]) return r[i] < r[j];
13     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
14 }
15 
16 void get_SA()
17 {
18     for(int i = 0; i <= n; i++)
19     {
20         SA[i] = i;
21         r[i] = i < n ? s[i] : -1;
22     }
23     for(k = 1; k <= n; k <<= 1)
24     {
25         sort(SA, SA + n + 1, cmp);
26         tmp[SA[0]] = 0;
27         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
28         memcpy(r, tmp, sizeof(r));
29     }
30     return;
31 }
32 
33 void get_height()
34 {
35     for(int i = 0; i <= n; i++) r[SA[i]] = i;
36     int k = 0;
37     for(int i = 0; i < n; i++)
38     {
39         if(k) k--;
40         int j = SA[r[i]-1];
41         while(s[i+k] == s[j+k]) k++;
42         h[r[i]] = k;
43     }
44     return;
45 }
46 
47 int main(void)
48 {
49     int T;
50     scanf("%d", &T);
51     for(int kase = 1; kase <= T; kase++)
52     {
53         scanf("%s", s);
54         n = strlen(s);
55         get_SA();
56         get_height();
57         int ans = n - SA[0];
58         for(int i = 1; i <= n; i++) ans += n - SA[i] - h[i];
59         printf("%d
", ans);
60     }
61     return 0;
62 }
    Aguin

    1.18

    什么都没干。

    1.19

    POJ 3693 Maximum repetition substring

    看了好多题解才看懂。然而并不好想阿。

      1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 const int maxn = 111111;
  7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn], rmq[maxn][20];
  8 char s[maxn];
  9 
 10 bool cmp(int i, int j)
 11 {
 12     if(r[i] != r[j]) return r[i] < r[j];
 13     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
 14 }
 15 
 16 void get_SA()
 17 {
 18     for(int i = 0; i <= n; i++)
 19     {
 20         SA[i] = i;
 21         r[i] = i < n ? s[i] : -1;
 22     }
 23     for(k = 1; k <= n; k <<= 1)
 24     {
 25         sort(SA, SA + n + 1, cmp);
 26         tmp[SA[0]] = 0;
 27         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
 28         memcpy(r, tmp, sizeof(r));
 29     }
 30     return;
 31 }
 32 
 33 void get_height()
 34 {
 35     for(int i = 0; i <= n; i++) r[SA[i]] = i;
 36     int k = 0;
 37     for(int i = 0; i < n; i++)
 38     {
 39         if(k) k--;
 40         int j = SA[r[i]-1];
 41         while(s[i+k] == s[j+k]) k++;
 42         h[r[i]] = k;
 43     }
 44     return;
 45 }
 46 
 47 void RMQ_init()
 48 {
 49     for(int i = 1; i <= n; i++) rmq[i][0] = h[i];
 50     for(int j = 1; (1 << j) <= n; j++)
 51         for(int i = 1; i + ( 1 << j ) - 1 <= n; i++)
 52             rmq[i][j] = min(rmq[i][j-1] , rmq[i+(1<<j-1)][j-1]);
 53 }
 54 
 55 int RMQ_query(int l, int r)
 56 {
 57     int k = 0;
 58     while( ( 1 << (k + 1) ) <= r - l + 1 ) k++;
 59     return min(rmq[l][k], rmq[r-(1<<k)+1][k]);
 60 }
 61 
 62 int lcp(int i, int j)
 63 {
 64     i = r[i], j = r[j];
 65     if(i > j) swap(i, j);
 66     return RMQ_query(i + 1, j);
 67 }
 68 
 69 int main(void)
 70 {
 71     int kase = 0;
 72     while(~scanf("%s", s) && s[0] != '#')
 73     {
 74         n = strlen(s);
 75         get_SA();
 76         get_height();
 77         RMQ_init();
 78         int Max = 1, MaxL = 1;
 79         for(int L = 1; L < n; L++)
 80         {
 81             for(int i = 0; i + L < n; i += L)
 82             {
 83                 int cur = lcp(i, i + L), tmp = cur / L + 1;
 84                 int pos = i - L + cur % L;
 85                 if(pos >= 0 && lcp(pos, pos + L) >= tmp * L) tmp++;
 86                 if(tmp > Max) Max = tmp, MaxL = L;
 87             }
 88         }
 89         for(int i = 1; i <= n; i++)
 90         {
 91             if(SA[i] + MaxL <= n && lcp(SA[i], SA[i] + MaxL) >= Max  * MaxL - MaxL)
 92             {
 93                 printf("Case %d: ", ++kase);
 94                 for(int j = 0; j < MaxL * Max; j++) putchar(s[SA[i]+j]);
 95                 puts(""); break;
 96             }
 97         }
 98     }
 99     return 0;
100 }
    Aguin

    POJ 3415 Common Substrings

    好烦啊。还要搞个单调栈。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long LL;
 7 const int maxn = 222222;
 8 int n, m, k, K, SA[maxn], r[maxn], tmp[maxn], h[maxn];
 9 LL sta[maxn], cnt[maxn];
10 char s[maxn];
11 
12 bool cmp(int i, int j)
13 {
14     if(r[i] != r[j]) return r[i] < r[j];
15     return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 );
16 }
17 
18 void get_SA()
19 {
20     for(int i = 0; i <= n; i++)
21     {
22         SA[i] = i;
23         r[i] = i < n ? s[i] : -1;
24     }
25     for(k = 1; k <= n; k <<= 1)
26     {
27         sort(SA, SA + n + 1, cmp);
28         tmp[SA[0]] = 0;
29         for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]);
30         memcpy(r, tmp, sizeof(r));
31     }
32     return;
33 }
34 
35 void get_height()
36 {
37     for(int i = 0; i <= n; i++) r[SA[i]] = i;
38     int k = 0;
39     for(int i = 0; i < n; i++)
40     {
41         if(k) k--;
42         int j = SA[r[i]-1];
43         while(s[i+k] == s[j+k]) k++;
44         h[r[i]] = k;
45     }
46     return;
47 }
48 
49 LL cal(int flag)
50 {
51     LL ret = 0, top = 0, sum = 0;
52     for(int i = 1; i <= n; i++)
53     {
54         if(h[i] < K) top = sum = 0;
55         else
56         {
57             int tmp = 0;
58             if( (SA[i-1] < m) ^ flag ) tmp = 1, sum += (LL) h[i] - K + 1;
59             while(top && sta[top-1] > h[i])
60             {
61                 top--;
62                 sum -= cnt[top] * (sta[top] - h[i]);
63                 tmp += cnt[top];
64             }
65             sta[top] = h[i], cnt[top++] = tmp;
66             if( (SA[i] > m) ^ flag ) ret += sum;
67         }
68     }
69     return ret;
70 }
71 
72 int main(void)
73 {
74     while(~scanf("%d", &K) && K)
75     {
76         scanf("%s", s);
77         m = strlen(s);
78         s[m] = '~';
79         scanf("%s", s + m + 1);
80         n = strlen(s);
81         get_SA();
82         get_height();
83         printf("%I64d
",cal(0) + cal(1));
84     }
85     return 0;
86 }
    Aguin

    1.20

    打个TC。没上黄。

    TC SRM 679 DIV2 500 ContestScoreboard

    FST了。就是一个模拟题。

    感觉学习到了就是sstream这个玩意儿。- -

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <sstream>
 6 #include <vector>
 7 #include <algorithm>
 8 using namespace std;
 9 typedef long long LL;
10 typedef pair<int, int> pii;
11 
12 struct team
13 {
14     string Name;
15     int id;
16     LL Score;
17 }T[111];
18 
19 bool cmp(team A, team B)
20 {
21     if(A.Score != B.Score) return A.Score > B.Score;
22     return A.Name < B.Name;
23 }
24 
25 class ContestScoreboard
26 {
27 public:
28     vector <int> findWinner(vector <string> scores)
29     {
30         int N = scores.size();
31         vector<pii> tmp[111];
32         vector<int> time;
33         int ans[111];
34         memset(ans, 0, sizeof(ans));
35         for(int i = 0; i < N; i++)
36         {
37             char c;
38             T[i].id = i;
39             stringstream ss;
40             ss.str(scores[i]);
41             ss >> T[i].Name;
42             int Sc, Ti;
43             while( ss >> Sc >> c >> Ti) tmp[i].push_back(pii(Ti,Sc)), time.push_back(Ti + 1);
44             sort(tmp[i].begin(), tmp[i].end());
45         }
46         time.push_back(0);
47         sort(time.begin(), time.end());
48         for(int i = 0; i < time.size(); i++)
49         {
50             for(int j = 0; j < N; j++)
51             {
52                 T[j].Score = 0LL;
53                 for(int k = 0; k < tmp[T[j].id].size(); k++)
54                 {
55                     if(tmp[T[j].id][k].first >= time[i]) break;
56                     T[j].Score += (LL) tmp[T[j].id][k].second;
57                 }
58             }
59             sort(T, T + N, cmp);
60             ans[T[0].id] = 1;
61         }
62         vector<int> ret;
63         for(int i = 0; i < N; i++) ret.push_back(ans[i]);
64         return ret;
65     }
66 };
    Aguin

    1.21

    TC SRM 679 DIV2 1000 ForbiddenStreets

    暴力思路很正确阿。然而没写完。

    Floyd时每两点维护一个边集。又一次用了SLT的集合运算。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 #include <set>
 6 #include <algorithm>
 7 using namespace std;
 8 int G[111][111];
 9 
10 class ForbiddenStreets
11 {
12 public:
13     vector <int> find(int N, vector <int> A, vector <int> B, vector <int> time)
14     {
15         int m = A.size();
16         memset(G, 0 ,sizeof(G));
17         set<int> r[111][111];
18         for(int i = 0; i < m; i++)
19         {
20             int a = A[i], b = B[i], d = time[i];
21             G[a][b] = G[b][a] = d;
22             r[a][b].insert(i);
23             r[b][a].insert(i);
24         }
25         for(int k = 0; k < N; k++)
26         {
27             for(int i = 0; i < N; i++)
28             {
29                 for(int j = 0; j < N; j++)
30                 {
31                     if(!G[i][k] || !G[k][j]) continue;
32                     if(!G[i][j] || G[i][k] + G[k][j] < G[i][j])
33                     {
34                         G[i][j] = G[i][k] + G[k][j];
35                         r[i][j].clear();
36                         set_union(r[i][k].begin(), r[i][k].end(), r[k][j].begin(), r[k][j].end(), inserter(r[i][j], r[i][j].begin()));
37                     }
38                     else if(G[i][k] + G[k][j] == G[i][j])
39                     {
40                         set<int> a, b;
41                         set_union(r[i][k].begin(), r[i][k].end(), r[k][j].begin(), r[k][j].end(), inserter(a, a.begin()));
42                         set_intersection(a.begin(), a.end(), r[i][j].begin(), r[i][j].end(), inserter(b, b.begin()));
43                         r[i][j] = b;
44                     }
45                 }
46             }
47         }
48         int cnt[2222];
49         memset(cnt, 0, sizeof(cnt));
50         for(int i = 0; i < N; i++)
51             for(int j = i + 1; j < N; j++)
52                 for(set<int>::iterator it = r[i][j].begin(); it != r[i][j].end(); it++) cnt[*it]++;
53         vector<int> ret;
54         for(int i = 0; i < m; i++) ret.push_back(cnt[i]);
55         return ret;
56     }
57 };
    Aguin

    1.22

    CF 620 D Professor GukiZ and Two Arrays

    场上比较sb。一直在贪。

    swap2次的话。把a的组合丢map里，再遍历b的组合去二分a。

    麻烦一点就是二分要检查两边的值，比如用lower_bound()要检查it和it--的key。

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <map>
 6 using namespace std;
 7 typedef long long LL;
 8 typedef pair<int, int> pii;
 9 LL a[2222], b[2222];
10 map<LL, pii> M;
11 map<LL, pii>::iterator it;
12 
13 int main(void)
14 {
15     int n, m;
16     LL sa = 0, sb = 0;
17     scanf("%d", &n);
18     for(int i = 1; i <= n; i++) scanf("%I64d", a + i), sa += a[i];
19     scanf("%d", &m);
20     for(int i = 1; i <= m; i++) scanf("%I64d", b + i), sb += b[i];
21     LL v = abs(sa - sb), pa = 0, pb = 0;
22     for(int i = 1; i <= n; i++)
23         for(int j = 1; j <= m; j++)
24             if(abs(sa - a[i] * 2LL - sb + b[j] * 2LL) < v)
25                 v = abs(sa - a[i] * 2LL - sb + b[j] * 2LL), pa = i, pb = j;
26     if(!pa) {printf("%I64d
0
", v); return 0;}
27 
28     for(int i = 1; i <= n; i++)
29         for(int j = i + 1; j <= n; j++)
30             M[(a[i] + a[j]) * 2LL] = pii(i, j);
31 
32     LL ppa = 0, ppb = 0;
33 
34     for(int i = 1; i <= m; i++)
35     {
36         for(int j = i + 1; j <= m; j++)
37         {
38             LL x = sa - sb + b[i] + b[i] + b[j] + b[j];
39             it = M.lower_bound(x);
40             if(it != M.end() && abs((*it).first - x) < v)
41             {
42                 v = abs((*it).first - x);
43                 pii tmp = (*it).second;
44                 pa = tmp.first, pb = i, ppa = tmp.second, ppb = j;
45             }
46             if(it != M.begin())
47             {
48                 it--;
49                 if(abs((*it).first - x) >= v) continue;
50                 v = abs((*it).first - x);
51                 pii tmp = (*it).second;
52                 pa = tmp.first, pb = i, ppa = tmp.second, ppb = j;
53             }
54         }
55     }
56 
57     if(!ppa) printf("%I64d
1
%I64d %I64d
", v, pa, pb);
58     else printf("%I64d
2
%I64d %I64d
%I64d %I64d
",v, pa, pb, ppa, ppb);
59     return 0;
60 }
    Aguin

    CF 620 E New Year Tree

    哭了太久没写不会写线段树了。

      1 #include <iostream>
  2 #include <cstdio>
  3 #define lson (2 * p)
  4 #define rson (2 * p + 1)
  5 #define mid ( (tr[p].l + tr[p].r) / 2 )
  6 using namespace std;
  7 typedef long long LL;
  8 const int maxn = 4e5 + 10;
  9 int cnt , h[maxn];
 10 int timer, dfn[maxn][2];
 11 int id[maxn<<1], col[maxn];
 12 
 13 struct edge
 14 {
 15     int to, pre;
 16 } e[maxn<<1];
 17 
 18 void add(int from, int to)
 19 {
 20     cnt++;
 21     e[cnt].pre = h[from];
 22     e[cnt].to = to;
 23     h[from] = cnt;
 24 }
 25 
 26 void dfs(int p, int fa)
 27 {
 28     dfn[p][0] = ++timer;
 29     id[timer] = p;
 30     for(int i = h[p]; i; i = e[i].pre)
 31     {
 32         int to = e[i].to;
 33         if(to == fa) continue;
 34         dfs(to, p);
 35     }
 36     dfn[p][1] = ++timer;
 37 }
 38 
 39 struct seg_tree
 40 {
 41     int l, r, tag;
 42     LL c;
 43 }tr[maxn<<3];
 44 
 45 void pushdown(int p)
 46 {
 47     if(tr[p].tag) tr[p].c = 1LL << tr[p].tag;
 48     if(tr[p].tag && tr[p].l != tr[p].r) tr[lson].tag = tr[rson].tag = tr[p].tag;
 49     tr[p].tag = 0;
 50 }
 51 
 52 void pushup(int p)
 53 {
 54     pushdown(lson), pushdown(rson);
 55     tr[p].c = tr[lson].c | tr[rson].c;
 56 }
 57 
 58 void build(int p, int l, int r)
 59 {
 60     tr[p].l = l, tr[p].r = r, tr[p].tag = tr[p].c = 0LL;
 61     if(l < r) build(lson, l, mid), build(rson, mid + 1, r), pushup(p);
 62     else if(id[l]) tr[p].c = 1LL << col[id[l]];
 63 }
 64 
 65 void modify(int p, int l, int r, int c)
 66 {
 67     if(tr[p].l >= l && tr[p].r <= r) tr[p].tag = c;
 68     pushdown(p);
 69     if(tr[p].l >= l && tr[p].r <= r) return;
 70     if(l <= mid) modify(lson, l, r, c);
 71     if(r > mid) modify(rson, l, r, c);
 72     pushup(p);
 73 }
 74 
 75 LL query(int p, int l, int r)
 76 {
 77     pushdown(p);
 78     if(tr[p].l >= l && tr[p].r <= r) return tr[p].c;
 79     LL ret = 0LL;
 80     if(l <= mid) ret |= query(lson, l, r);
 81     if(r > mid) ret |= query(rson, l, r);
 82     pushup(p);
 83     return ret;
 84 }
 85 
 86 int main(void)
 87 {
 88     int n, m;
 89     scanf("%d %d", &n, &m);
 90     for(int i = 1; i <= n; i++) scanf("%d", col + i);
 91     for(int i = 1; i < n; i++)
 92     {
 93         int u, v;
 94         scanf("%d %d", &u, &v);
 95         add(u, v), add(v, u);
 96     }
 97     dfs(1, 0);
 98     build(1, 1, timer);
 99     for(int i = 1; i <= m; i++)
100     {
101         int op, v, c, ans = 0;
102         scanf("%d", &op);
103         if(op == 1)
104         {
105             scanf("%d %d", &v, &c);
106             modify(1, dfn[v][0], dfn[v][1], c);
107         }
108         else
109         {
110             scanf("%d", &v);
111             LL ret = query(1, dfn[v][0], dfn[v][1]);
112             while(ret) ans += ret & 1LL, ret >>= 1LL;
113             printf("%d
", ans);
114         }
115     }
116     return 0;
117 }
    Aguin

    1.23

    HDU 5611 Baby Ming and phone number

    SBSBSBSBSBSB

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 typedef long long LL;
 6 int d[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
 7 char * num[22] = {
 8     "00000", "11111", "22222", "33333", "44444", "55555", "66666", "77777", "88888", "99999",
 9     "01234", "12345", "23456", "34567", "45678", "56789",
10     "98765", "87654", "76543", "65432", "54321", "43210"
11 };
12 
13 bool judge(char * s)
14 {
15     for(int i = 0; i < 22; i++) if(strcmp(num[i], s + 6) == 0) return true;
16     int y = ( s[3] - '0' ) * 1000 + ( s[4] - '0' ) * 100 + ( s[5] - '0' ) * 10 + ( s[6] - '0' );
17     int m = ( s[7] - '0' ) * 10 + ( s[8] - '0' );
18     int da = ( s[9] - '0' ) * 10 + ( s[10] - '0' );
19     if(y < 1980 || y > 2016) return false;
20     int leap = 0;
21     if( ( (y % 4 == 0) && (y % 100 != 0) ) || (y % 400 == 0) ) leap = 1;
22     if(m < 1 || m > 12) return false;
23     if(da >= 1 && da <= d[m] + (m == 2 && leap) ) return true;
24     return false;
25 }
26 
27 int main(void)
28 {
29     int T;
30     scanf("%d", &T);
31     while(T--)
32     {
33         LL n, a, b, ans = 0;
34         scanf("%I64d%I64d%I64d", &n, &a, &b);
35         for(int i = 0; i < n; i++)
36         {
37             char ss[20];
38             scanf("%s", ss);
39             if(judge(ss)) ans += a;
40             else ans += b;
41         }
42         printf("%I64d
", ans);
43     }
44     return 0;
45 }
    Aguin
  • 相关阅读:
    【Beta版本】冲刺-Day6
    【Beta版本】冲刺-Day5
    【Beta版本】冲刺-Day4
    【Beta版本】冲刺-Day3
    【Beta版本】冲刺-Day2
    【Beta版本】冲刺-Day1
    内存管理和虚拟内存
    C++11中的智能指针和锁的类型以及它们背后的RAII机制
    操作系统中面试时问到的相关知识点
    网络层（IP）相关知识
  • 原文地址：https://www.cnblogs.com/Aguin/p/5136656.html
Copyright © 2011-2022 走看看