EOJ1054

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int n, C;
 5 int nc, nw;
 6 int ans;
 7 int cost[60][3], weight[60][3];
 8 void dfs(int i) {
 9     if (i == n) ans= min(ans, nw);
10     for (int j = 0; j < 3; j++) {
11         nc += cost[i][j];
12         nw += weight[i][j];
13         if (nc < C&&nw<ans) 
14             dfs(i + 1);
15         nc -= cost[i][j];
16         nw -= weight[i][j];
17     }
18 }
19 int main() {
20     while (cin >> n >> C) {
21         for (int i = 0; i < n; i++) {
22             for (int j = 0; j < 3; j++) {
23                 cin >> weight[i][j] >> cost[i][j];
24             }
25         }
26         ans = 0x3f3f3f3f;
27         dfs(0);
28         cout << ans<<endl;
29     }
30     return 0;
31 }

View Code

题意：输入两个数n，C表示以下n行，之后每行6个数，两个数一组，第一个数表示该部件质量，第二个数表示该部件的花费。要求在总cost<C的条件下，从每行选一个部件，使得总质量最小。

思路：

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int n, C;
 5 int nc, nw;
 6 int ans;
 7 int cost[60][3], weight[60][3];
 8 void dfs(int i) {
 9     if (i == n) ans= min(ans, nw);
10     for (int j = 0; j < 3; j++) {
11         nc += cost[i][j];
12         nw += weight[i][j];
13         if (nc < C&&nw<ans) 
14             dfs(i + 1);
15         nc -= cost[i][j];
16         nw -= weight[i][j];
17     }
18 }
19 int main() {
20     while (cin >> n >> C) {
21         for (int i = 0; i < n; i++) {
22             for (int j = 0; j < 3; j++) {
23                 cin >> weight[i][j] >> cost[i][j];
24             }
25         }
26         ans = 0x3f3f3f3f;
27         dfs(0);
28         cout << ans<<endl;
29     }
30     return 0;
31 }

View Code