[I will also give a brief description of the problem] Software Company

[Description]

N people, two projects, each having M same parts. The minimum time required for each person to complete a certain project is different.

[Input format]

Row 1: Number N and part M.

Next n rows, two numbers per line: the time required for I to complete one part of the two projects.

[Output format]

One row: the fastest time.

[Example input]

3 20

1 1

2 4

1 6

[Sample output]

18

[Data Scope]

1 <= n, m <= 100.

[Analysis]

The total time used by the binary enumeration, and then the dynamic plan is used to determine whether the result can be completed. Similar to the question "Green Channel" issued by students in Tangshan No. 1 school. But it is simpler than that.

Dynamic Identification:

The current time to be determined is mid. f [I] [J] is used to represent the previous I, and the first project of J is completed. The second project can complete up to several copies. The state transition equation is f [I] [J] = max (F [I-1] [J-K] + (mid-K * W [I] [0]). /W [I] [1]). Finally, determine whether f [N] [m] is greater than M.

// Problem: software company // By Sephiroth Lee/Date: 10/9/2010 # include <stdio. h> # include <string. h ># include <iostream> # define maxn 110 using namespace STD; int W [maxn] [2], F [maxn] [maxn]; int n, m, max_time, l, R, mid; bool check (INT mid) {memset (F,-1, sizeof (f); F [0] [0] = 0; for (INT I = 1; I <= N; ++ I) for (Int J = 0; j <= m; ++ J) {If (F [I-1] [J] <0) continue; For (int K = 0; (j + k <= m) & (mid> = W [I] [0] * k); ++ K) if (F [I] [J + k] <F [I-1] [J] + (mid-W [I] [0] * K) /W [I] [1]) f [I] [J + k] = f [I-1] [J] + (mid-W [I] [0] * K) /W [I] [1];} return f [N] [m]> = m;} int main () {freopen ("software. in "," r ", stdin); freopen (" software. out "," W ", stdout); scanf (" % d ", & N, & M); For (INT I = 1; I <= N; ++ I) {scanf ("% d", & W [I] [0], & W [I] [1]); if (W [I] [0]> max_time) max_time = W [I] [0]; If (W [I] [1]> max_time) max_time = W [I] [1];} r = max_time * m; L = 1; while (L <= r) {mid = (L + r)/2; if (check (MID) r = mid-1; else l = Mid + 1;} printf ("% d \ n", L); Return 0 ;}