wenbao与最优比率生成树

 1 #include "iostream"
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <stdio.h>
 5 using namespace std;
 6 
 7 #define ll long long
 8 #define eps 1e-8
 9 const int maxn = 1009;
10 double a[maxn], b[maxn], d[maxn];
11 int n, k;
12 
13 double cheak(double mid) {
14     for(int i = 0; i < n; ++i) d[i] = a[i]-mid*b[i];
15     sort(d, d+n);
16     double sum = 0;
17     for(int i = 0; i < k; ++i) sum += d[n-1-i];
18     return sum;
19 }
20 
21 int main() {
22 #ifdef wenbao
23     freopen("in", "r", stdin);
24 #endif
25     while(~scanf("%d%d", &n, &k)) {
26         if(n == 0 && k == 0) break;
27         k = n-k;
28         for(int i = 0; i < n; ++i) scanf("%lf", &a[i]);
29         for(int i = 0; i < n; ++i) scanf("%lf", &b[i]);
30         double l = 0.0, r = 1000000000.0, mid;
31         while(fabs(r-l) > eps) {
32             mid = (r+l)/2.0;
33             if(cheak(mid) >= 0) l = mid;
34             else r = mid;
35         }
36         printf("%.0lf\n", floor(l*100+0.5));
37     }
38     return 0;
39 }

 1 #include "iostream"
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <stdio.h>
 5 using namespace std;
 6 
 7 #define ll long long
 8 #define eps 1e-8
 9 const int maxn = 1009;
10 double a[maxn], b[maxn], d[maxn];
11 int id[maxn], n, k;
12 
13 bool cmp(int x, int y) {
14     return d[x] > d[y];
15 }
16 
17 double cheak(double l) {
18     for(int i = 0; i < n; ++i) d[i] = a[i]-l*b[i], id[i] = i;
19     sort(id, id+n, cmp);
20     double sum1 = 0.0, sum2 = 0.0;
21     for(int i = 0; i < n-k; ++i) sum1 += a[id[i]], sum2 += b[id[i]];
22     return sum1/sum2;
23 }
24 
25 int main() {
26 #ifdef wenbao
27     freopen("in", "r", stdin);
28 #endif
29     while(~scanf("%d%d", &n, &k)) {
30         if(n == 0 && k == 0) break;
31         for(int i = 0; i < n; ++i) scanf("%lf", &a[i]);
32         for(int i = 0; i < n; ++i) scanf("%lf", &b[i]);
33         double l = 0.0, r;
34         while(1) {
35             r = cheak(l);
36             if(fabs(r-l) < eps) break;
37             l = r;
38         }
39         printf("%.0lf\n", floor(l*100+0.5));
40     }
41     return 0;
42 }

 1 #include "iostream"
 2 #include <queue>
 3 #include <string.h>
 4 #include <cmath>
 5 #include <stdio.h>
 6 using namespace std;
 7 
 8 #define eps 1e-4
 9 const int maxn = 1009;
10 int n;
11 double a[maxn], b[maxn], c[maxn], cost[maxn][maxn], di[maxn][maxn], dist[maxn];
12 
13 double prim(double x) {
14     double ans = 0.0;
15     for(int i = 2; i <= n; ++i){
16         dist[i] = cost[1][i] - x*di[1][i];
17     }
18     dist[1] = -1;
19     int k;
20     for(int i = 2; i <= n; ++i){
21         double mi = 10000000000000.0;
22         for(int j = 1; j <= n; ++j){
23             if(dist[j] != -1 && dist[j] < mi){
24                 mi = dist[j];
25                 k = j;
26             }
27         }
28         dist[k] = -1;
29         ans += mi;
30         for(int j = 1; j <= n; ++j){
31             if(dist[j] != -1 && dist[j] > cost[k][j] - x*di[k][j]){
32                 dist[j] = cost[k][j] - x*di[k][j];
33             }
34         }
35     }
36     return ans;
37 }
38 
39 int main(){
40 #ifdef wenbao
41     freopen("in", "r", stdin);
42 #endif
43     while(~scanf("%d", &n) && n){
44         for(int i = 1; i <= n; ++i){
45             scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
46             for(int j = 1; j < i; ++j){
47                 double x = (a[i]-a[j])*(a[i]-a[j]) + (b[i]-b[j])*(b[i]-b[j]);
48                 cost[j][i] = cost[i][j] = (c[i] > c[j]) ? c[i] - c[j] : c[j] - c[i];
49                 di[i][j] = di[j][i] = sqrt(x);
50             }
51         }
52         double l = 0.0, r = 100000.0, mid;
53         while((r-l) > eps){
54             //cout<<l<<"&&"<<r<<endl;
55             mid = (l+r)/2.0;
56             if(prim(mid) >= 0){
57                 l = mid;
58             }else{
59                 r = mid;
60             }
61         }
62         printf("%.3lf\n", l);
63     }
64     return 0;
65 }

 1 #include "iostream"
 2 #include <queue>
 3 #include <string.h>
 4 #include <cmath>
 5 #include <stdio.h>
 6 using namespace std;
 7 
 8 #define eps 1e-4
 9 const int maxn = 1009;
10 int n, id[maxn];
11 double a[maxn], b[maxn], c[maxn], cost[maxn][maxn], di[maxn][maxn], dist[maxn];
12 bool vis[maxn];
13 
14 
15 double prim(double x) { 
16     memset(vis, false, sizeof(vis));
17     for(int i = 2; i <= n; ++i) dist[i] = cost[1][i] - di[1][i]*x, id[i] = 1;
18     vis[1] = true, dist[1] = 0;
19     int k;
20     double sum1 = 0.0, sum2 = 0.0;
21     for(int i = 2; i <= n; ++i){
22         double mi = 100000000000.0;
23         for(int j = 2; j <= n; ++j){
24             if(!vis[j] && dist[j] < mi){
25                 k = j, mi = dist[j];
26             }
27         }
28         vis[k] = true;
29         sum1 += cost[id[k]][k], sum2 += di[id[k]][k];
30         for(int j = 2; j <= n; ++j){
31             if(!vis[j] && dist[j] > cost[k][j] - di[k][j]*x){
32                 dist[j] = cost[k][j] - di[k][j]*x;
33                 id[j] = k;
34             }
35         }
36     }
37     return sum1/sum2;
38 }
39 
40 int main(){
41 #ifdef wenbao
42     freopen("in", "r", stdin);
43 #endif
44     while(~scanf("%d", &n) && n){
45         for(int i = 1; i <= n; ++i){
46             scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
47             for(int j = 1; j < i; ++j){
48                 double x = (a[i]-a[j])*(a[i]-a[j]) + (b[i]-b[j])*(b[i]-b[j]);
49                 cost[j][i] = cost[i][j] = (c[i] > c[j]) ? c[i] - c[j] : c[j] - c[i];
50                 di[i][j] = di[j][i] = sqrt(x);
51             }
52         }
53         double l = 0.0, r;
54         while(1){
55             r = prim(l);
56             if(fabs(r-l) < eps){
57                 break;
58             }else{
59                 l = r;
60             }
61         }
62         printf("%.3lf\n", l);
63     }
64     return 0;
65 }