UVA 11269 - Setting Problems（贪心）
Problem J
Setting Problems
Input: Standard Input
Output: Standard Output
As you see, setting problems for a programming contest is a tough job. There are so many things to do like creating problems, solutions, data files, verification of problem statements, writing alternate judge solutions etc. etc. Given the responsibility of creating the problemset for ‘If you can not crash judges by solutions, crash contestants by problems’ programming contest, Sultan & GolapiBaba have realized that to the backbone. Finally they agree that they will set N problems for the contest. For each of the problems, first Sultan will create the problem statement, solution & i/o data. After he finishes his work, GolapiBaba does the verification & alternate solution writing part for that particular problem. Each of them needs a particular amount of time to complete their tasks for a certain problem. Also, none of them works on more than one problem at a time. Note that, GolapiBaba can start working on a problem immediately after Sultan finishes that problem or he may wish to start that problem later.
You will be given the times that Sultan & GolapiBaba requires to complete their respective tasks for every single problem. Determine the minimum possible time required to complete the whole problemset.
Input
There are around 50 test cases. Each test case starts with a single integer N (1<=N<=20), the number of problems in the contest. The next line contains N integers Si (1<=Si<=100, 1<=i<=N) where Si denotes the time required for Sultan to complete his tasks for problem i. The next line has N more integers Gi (1<=Gi<=100, 1<=i<=N) where Gi denotes the time required for Golapibaba to complete his tasks on problem i.
Output
For each test case, print the minimum time required to complete the problemset.
| Sample Input |
Sample Output |
| 3 8 1 6 1 6 3 3 4 5 6 1 1 6
|
16 16
|
题意：有n个任务，S需要Si时间，G需要G时间，每个任务要S完成了G才能去做。求最少完成任务时间。
思路：贪心，对于两个任务，只要比较谁先做花的时间少即可。时间分别为(g1 - s2 < 0 ? 0 : g1 - s2) + g2 和(g2 - s1 < 0 ? 0 : g2 - s1) + g1; 小的优先完成。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 105;
int n;
struct Work {
int s, g;
} w[N];
bool cmp(Work a, Work b) {
return (a.g - b.s < 0 ? 0 : a.g - b.s) + b.g < (b.g - a.s < 0 ? 0 : b.g - a.s) + a.g;
}
void init() {
for (int i = 0; i < n; i++)
scanf("%d", &w[i].s);
for (int j = 0; j < n; j++)
scanf("%d", &w[j].g);
}
int solve() {
sort(w, w + n, cmp);
int have = 0, time = 0;
for (int i = 0; i < n; i++) {
have -= w[i].s;
if (have < 0) have = 0;
time += w[i].s;
have += w[i].g;
}
return time + have;
}
int main() {
while (~scanf("%d", &n)) {
init();
printf("%d\n", solve());
}
return 0;
}