比赛-Round (11 Jul)

1. 小区划分

动规。题目略坑，通过样例发现，每个住户必须属于一个小区。

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 double f[810][85], A[810], B[810];
 8 
 9 int main()
10 {
11     int i, j, k, N, K;
12     scanf("%d%d", &N, &K);
13     for (i = 1; i <= N; ++i)
14         scanf("%lf", &A[i]), A[i] += A[i-1];
15     for (i = 1; i <= N; ++i)
16         scanf("%lf", &B[i]), B[i] += B[i-1];
17     for (i = 1; i <= N; ++i)
18         for (k = 1; k <= i && k <= K; ++k) {
19             if (k == 1) { f[i][1] = fabs(A[i]-B[i]); continue; }
20             for (j = k-1; j < i; ++j) {
21                 f[i][k] = max(f[i][k], f[j][k-1]+fabs((A[i]-A[j])-(B[i]-B[j])));
22             }
23         }
24     printf("%.6lf\n", f[N][K]);
25     return 0;
26 }

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2. 直线的交点

把各直线与两条平行线的交点处理一下就转换成了逆序对问题。看代码比较容易理解。

 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 
 6 const int _N = 120000;
 7 
 8 typedef long long LL;
 9 
10 double K[_N], B[_N], X[_N], Y[_N], SX[_N], SY[_N], C[_N], GG[_N], k, a, b;
11 int N;
12 
13 void Add(int k, int d)
14 {
15     for (int i = k; i <= N; i += i & -i)
16         C[i] += d;
17     return;
18 }
19 
20 LL GetSum(int k)
21 {
22     LL tmp = 0;
23     for (int i = k; i; i ^= i & -i)
24         tmp += C[i];
25     return tmp;
26 }
27 
28 int main()
29 {
30     int i;
31     scanf("%lf%lf%lf%d", &k, &a, &b, &N);
32     
33     for (i = 1; i <= N; ++i) {
34         scanf("%lf%lf", &K[i], &B[i]);
35         X[i] = (B[i]-a)/(k-K[i]);
36         SX[i] = X[i];
37         Y[i] = (B[i]-b)/(k-K[i]);
38         SY[i] = Y[i]; 
39     }
40     
41     sort(SX+1, SX+1+N);
42     sort(SY+1, SY+1+N);
43     
44     for (i = 1; i <= N; ++i) {
45         LL tmpx = (LL)(lower_bound(SX+1, SX+1+N, X[i])-SX);
46         LL tmpy = (LL)(lower_bound(SY+1, SY+1+N, Y[i])-SY);
47         GG[tmpx] = tmpy;
48     }
49     
50     
51     LL ggans = 0;
52     for (i = 1; i <= N; ++i) {
53         ggans += GetSum(N)-GetSum(GG[i]);
54         Add(GG[i], 1);
55     }
56     printf("%lld\n", ggans);
57     
58     return 0;
59 }

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3. 数三角形

各种计数推，只要能搞出来就好了。

 1 #include <cstdio> 
 2 
 3 typedef long long LL;
 4 
 5 LL cnt[120000][2];
 6 
 7 LL C_k_2(LL v) { return v >= 2 ? v*(v-1)/2 : 0; }
 8 
 9 int main()
10 {
11     LL N, M, i;
12     scanf("%lld%lld", &N, &M);
13     for (i = 1; i <= M; ++i) {
14         LL x, y, w;
15         scanf("%lld%lld%lld", &x, &y, &w);
16         ++cnt[x][w-1], ++cnt[y][w-1];
17     }
18     LL X = 0, Y = 0;
19     for (i = 1; i <= N; ++i) {
20         LL cnt1 = cnt[i][0], cnt2 = cnt[i][1];
21         LL cnt3 = N-1-cnt[i][0]-cnt[i][1];
22         X += cnt1*cnt2+cnt2*cnt3+cnt1*cnt3;
23         Y += C_k_2(cnt1)+C_k_2(cnt2)+C_k_2(cnt3);
24     }
25     printf("%lld\n", X-2*Y);
26     
27     return 0;
28 }

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