Problem C: Work Reduction
Paperwork is beginning to pile up on your desk, and tensions at the workplace are starting to mount. Your boss has threatened to fire you if you don't make any progress by the end of the day. You currently have N units of paperwork on your desk, and your boss demands that you have exactly M units of paperwork left by the end of the day.
The only hope for you now is to hire help. There are various agencies which offer paperwork reduction plans:
For $A they will reduce your paperwork by one unit.
For $B they will reduce your entire paperwork by half (rounding down when necessary).
Note that work can never be reduced to less than 0.
Your task now is to produce a sorted table of agency names and their respective minimum costs to solve your workload problem.
The first line of input consists of a single positive integer representing the number of cases to follow. Each case begins with three positive integers separated by spaces: N - your starting workload, M - your target workload, and L - the number of work reduction agencies available to you, (1 <= M <= N <= 100000, 1 <= L <= 100). The next L lines have the format "[agency name]:A,B", where A and B are the rates as described above for the given agency. (0 <= A,B <= 10000) The length of the agency name will be between 1 and 16, and will consist only of capital letters. Agency names will be unique.
For each test case, print "Case X", with X being the case number, on a single line, followed by the table of agency names and their respective minimum costs, sorted in non-decreasing order of minimum costs. Sort job agencies with identical minimum costs in alphabetical order by agency name. For each line of the table, print out the agency name, followed by a space, followed by the minimum required cost for that agency to solve your problem.
Sample Input
2
100 5 3
A:1,10
B:2,5
C:3,1
1123 1122 5
B:50,300
A:1,1000
C:10,10
D:1,50
E:0,0
Sample Output
Case 1
C 7
B 22
A 37
Case 2
E 0
A 1
D 1
C 10
B 50
題意：有n個任務，要求完成到m個。。現在有l家公司，每家公司有一個名字有A代表完成一項工作花A元，B代表完成一半的工作花B元。。。對於這個完成一半。題目中是說rounding down。。我開始以為是向下取整。結果答案跑出來都跟樣例不一樣。。後來才明白其實是四捨五入。。。還有題目中公司的名字是字符串。我一開始看樣例想當然以為是一個字符。結果WA了幾次。。。。。哎
還有一個注意點。輸出要按花費少的排序。就是如果花費相同，就按名字的字典序排序。
思路：貪心。。每次比較一件件完成和完成一半誰花得少。就用哪個方案。直到n / 2 >= m 為止。。剩下還沒完成的部分。只能一項項完成（因為最後剩下的任務一定要是m）。。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int t, n, m, l;
struct Work {
char name[20];
int a, b;
int value;
} w[105];
int cmp(Work aa, Work bb) {
if (aa.value != bb.value)
return aa.value < bb.value;
if (strcmp(aa.name, bb.name) < 0)
return 1;
else
return 0;
}
void solve(int nn) {
int num = n;
while (num / 2 >= m) {
if (w[nn].b < (num -num / 2) * w[nn].a)
w[nn].value += w[nn].b;
else
w[nn].value += w[nn].a * (num - num / 2);
num /= 2;
}
w[nn].value += (num - m) * w[nn].a;
}
int main() {
int Case = 1;
scanf("%d", &t);
while (t --) {
memset(w, 0, sizeof(w));
scanf("%d%d%d%*c", &n, &m, &l);
for (int i = 0; i < l; i ++) {
char sb; int j = 0;
while ((sb = getchar()) != EOF && sb != ':') {
w[i].name[j ++] = sb;
}
w[i].name[j] = '\0';
scanf("%d%*c%d%*c", &w[i].a, &w[i].b);
}
for (int i = 0; i < l; i ++) {
solve(i);
}
sort(w, w + l, cmp);
printf("Case %d\n", Case ++);
for (int i = 0; i < l; i ++)
printf("%s %d\n", w[i].name, w[i].value);
}
return 0;
}
