AtCoder Grand Contest 012 题解

A - AtCoder Group Contest

排序一下，贪心取就好了。

 1 //waz
 2 #include <bits/stdc++.h>
 3  
 4 using namespace std;
 5  
 6 #define mp make_pair
 7 #define pb push_back
 8 #define fi first
 9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12  
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18  
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22  
23 int F()
24 {
25     char ch;
26     int x, a;
27     while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
28     if (ch == '-') ch = getchar(), a = -1;
29     else a = 1;
30     x = ch - '0';
31     while (ch = getchar(), ch >= '0' && ch <= '9')
32         x = (x << 1) + (x << 3) + ch - '0';
33     return a * x;
34 }
35  
36 const int N = 300010;
37  
38 int n, a[N];
39  
40 int main()
41 {
42     gi(n);
43     for (int i = 1; i <= 3 * n; ++i) gi(a[i]);
44     sort(a + 1, a + 3 * n + 1);
45     int cnt = 0;
46     long long ans = 0;
47     for (int i = 3 * n - 1; i; i -= 2)
48     {
49         ans += a[i];
50         ++cnt;
51         if (cnt == n) break;
52     }
53     printf("%lld\n", ans);
54     return 0;
55 }

 

B - Splatter Painting

d很小，倒着暴力即可，每个点只会被最多遍历10次。

 1 //waz
 2 #include <bits/stdc++.h>
 3  
 4 using namespace std;
 5  
 6 #define mp make_pair
 7 #define pb push_back
 8 #define fi first
 9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12  
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18  
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22  
23 int F()
24 {
25     char ch;
26     int x, a;
27     while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
28     if (ch == '-') ch = getchar(), a = -1;
29     else a = 1;
30     x = ch - '0';
31     while (ch = getchar(), ch >= '0' && ch <= '9')
32         x = (x << 1) + (x << 3) + ch - '0';
33     return a * x;
34 }
35  
36 const int N = 1e5 + 10;
37  
38 int n, m;
39  
40 VI edge[N];
41  
42 int mxd[N], co[N];
43  
44 void dfs(int u, int d, int c)
45 {
46     if (d <= mxd[u]) return;
47     if (!co[u]) co[u] = c;
48     mxd[u] = d;
49     for (auto v : edge[u]) dfs(v, d - 1, c);
50 }
51  
52 int v[N], d[N], c[N];
53  
54 int main()
55 {
56     gii(n, m);
57     for (int i = 1; i <= n; ++i) mxd[i] = -1;
58     for (int i = 1; i <= m; ++i)
59     {
60         int u, v;
61         gii(u, v);
62         edge[u].pb(v);
63         edge[v].pb(u);
64     }
65     int q;
66     gi(q);
67     for (int i = 1; i <= q; ++i)
68         giii(v[i], d[i], c[i]);
69     for (int i = q; i; --i)
70         dfs(v[i], d[i], c[i]);
71     for (int i = 1; i <= n; ++i)
72         printf("%d\n", co[i]);
73 }

 

C - Tautonym Puzzle

倍增。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
long long n;
 
deque<int> l, r;
 
int tot;
 
void solve(long long n)
{
    if (n == 1) return;
    solve(n >> 1);
    l.push_front(++tot);
    r.push_front(tot);
    if (n & 1)
    {
        l.push_front(++tot);
        r.push_back(tot);
    }
}
 
int main()
{
    scanf("%lld", &n);
    solve(n + 1);
    printf("%d\n", l.size() + r.size());
    for (auto x : l) printf("%d ", x);
    for (auto x : r) printf("%d ", x);
    return 0;
}

 

D - Colorful Balls

我们找到那些可以自由移动的球，算一下方案数就好了。

  1 //waz
  2 #include <bits/stdc++.h>
  3  
  4 using namespace std;
  5  
  6 #define mp make_pair
  7 #define pb push_back
  8 #define fi first
  9 #define se second
 10 #define ALL(x) (x).begin(), (x).end()
 11 #define SZ(x) ((int)((x).size()))
 12  
 13 typedef pair<int, int> PII;
 14 typedef vector<int> VI;
 15 typedef long long int64;
 16 typedef unsigned int uint;
 17 typedef unsigned long long uint64;
 18  
 19 #define gi(x) ((x) = F())
 20 #define gii(x, y) (gi(x), gi(y))
 21 #define giii(x, y, z) (gii(x, y), gi(z))
 22  
 23 int F()
 24 {
 25     char ch;
 26     int x, a;
 27     while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
 28     if (ch == '-') ch = getchar(), a = -1;
 29     else a = 1;
 30     x = ch - '0';
 31     while (ch = getchar(), ch >= '0' && ch <= '9')
 32         x = (x << 1) + (x << 3) + ch - '0';
 33     return a * x;
 34 }
 35  
 36 const int N = 2e5 + 10;
 37  
 38 const int mod = 1e9 + 7;
 39  
 40 int n, x, y, c[N], w[N], mi[N], cnt[N];
 41  
 42 int fac[N], ifac[N];
 43  
 44 int C(int n, int m)
 45 {
 46     if (n < m || m < 0) return 0;
 47     return 1LL * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
 48 }
 49  
 50 int main()
 51 {
 52     giii(n, x, y);
 53     for (int i = 1; i <= n; ++i)
 54     {
 55         gii(c[i], w[i]);
 56         if (!mi[c[i]]) mi[c[i]] = i;
 57         else if (w[mi[c[i]]] > w[i]) mi[c[i]] = i;
 58         ++cnt[c[i]];
 59     }
 60     fac[0] = 1;
 61     for (int i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
 62     ifac[0] = ifac[1] = 1;
 63     for (int i = 2; i <= n; ++i) ifac[i] = 1LL * (mod - mod / i) * ifac[mod % i] % mod;
 64     for (int i = 2; i <= n; ++i) ifac[i] = 1LL * ifac[i - 1] * ifac[i] % mod;
 65     int cho = 0;
 66     for (int i = 1; i <= n; ++i)
 67     {
 68         if (!mi[i]) continue;
 69         if (!mi[cho]) cho = i;
 70         else
 71         {
 72             if (w[mi[cho]] > w[mi[i]])
 73                 cho = i;
 74         }
 75     }
 76     int t = 1e9;
 77     for (int i = 1; i <= n; ++i)
 78         if (mi[i] && cho != i) t = min(t, w[mi[i]]);
 79     for (int i = 1; i <= n; ++i)
 80     {
 81         if (cho != c[i])
 82         {
 83             if (i == mi[c[i]]) continue;
 84             if (w[i] + w[mi[c[i]]] <= x) continue;
 85             if (w[i] + w[mi[cho]] <= y) continue;
 86         }
 87         else
 88         {
 89             if (i == mi[c[i]]) continue;
 90             if (w[i] + w[mi[c[i]]] <= x) continue;
 91             if (w[i] + t <= y) continue;
 92         }
 93         //cerr << i << endl;
 94         --cnt[c[i]];
 95     }
 96     /*for (int i = 1; i <= n; ++i)
 97         cerr << cnt[i] << endl;*/
 98     int num = 0;
 99     //cerr << w[mi[2]] << endl;
100     for (int i = 1; i <= n; ++i)
101     {
102         if (mi[i] && w[mi[i]] + w[mi[cho]] <= y) num += cnt[i];
103     }
104     //cerr << num << endl;
105     int ans = 1;
106     for (int i = 1; i <= n; ++i)
107     {
108         if (mi[i] && w[mi[i]] + w[mi[cho]] <= y)
109         {
110             //cerr << cnt[i] << endl;
111             ans = 1LL * ans * C(num, cnt[i]) % mod;
112             num -= cnt[i];
113         }
114     }
115     printf("%d\n", ans);
116     return 0;
117 }

 

E - Camel and Oases

状压DP，算出不用v的f和g，然后判断一下每个区间的合法性即可。

 1 //waz
 2 #include <bits/stdc++.h>
 3  
 4 using namespace std;
 5  
 6 #define mp make_pair
 7 #define pb push_back
 8 #define fi first
 9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12  
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18  
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22  
23 int F()
24 {
25     char ch;
26     int x, a;
27     while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
28     if (ch == '-') ch = getchar(), a = -1;
29     else a = 1;
30     x = ch - '0';
31     while (ch = getchar(), ch >= '0' && ch <= '9')
32         x = (x << 1) + (x << 3) + ch - '0';
33     return a * x;
34 }
35  
36 const int N = 1 << 20;
37  
38 int n, v, x[200010];
39  
40 int l[22][200010], r[22][200010];
41  
42 int val[22], cnt, f[N], g[N];
43  
44 vector<PII> t;
45  
46 int main()
47 {
48     gii(n, v);
49     for (int i = 1; i <= n; ++i) gi(x[i]);
50     for (int i = v; ; i = i >> 1)
51     {
52         val[++cnt] = i;
53         for (int j = 1; j <= n; ++j)
54         {
55             int k = j;
56             while (k < n && x[k + 1] - x[k] <= i) ++k;
57             for (int p = j; p <= k; ++p) l[cnt][p] = j, r[cnt][p] = k;
58             if (cnt == 1) t.pb(mp(j, k));
59             j = k;
60         }
61         if (!i) break;
62     }
63     int all = cnt - 1;
64     for (int i = 0; i < (1 << all); ++i) g[i] = n + 1;
65     for (int i = 1; i < (1 << all); ++i)
66     {
67         for (int j = 0; j < all; ++j)
68             if (i & (1 << j))
69             {
70                 int t = f[i ^ (1 << j)];
71                 f[i] = max(f[i], r[j + 2][t + 1]);
72                 t = g[i ^ (1 << j)];
73                 g[i] = min(g[i], l[j + 2][t - 1]);
74             }
75     }
76     for (auto v : t)
77     {
78         bool flag = 0;
79         int s = (1 << all) - 1;
80         for (int i = 0; i < (1 << all); ++i)
81         {
82             if (f[i] >= v.fi - 1 && v.se + 1 >= g[s ^ i])
83                 flag = 1;
84         }
85         for (int i = v.fi; i <= v.se; ++i)
86             puts(flag ? "Possible" : "Impossible");
87     }
88 }

 

F - Prefix Median

不会，咕着qwq（最近好像已经咕了3题了）

posted @ 2018-09-18 22:06  AnzheWang  阅读(202)  评论(0)  编辑  收藏  举报