[UVA][] 10273 - Eat or Not to Eat＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2014-02-17 09:11:25| 人2,925| 回0 | 上一篇 | 下一篇

[UVA][] 10273 - Eat or Not to Eat

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Problem E
Eat or not to Eat?
Input: Standard Input
Output: Standard Output

A young farmer has N cows, but they produced really really a very very small amount of milk. John cannot live on the milk they made, so he's planning to eat some of the 'worst' cows to get rid of hunger. Each day, John chooses the cow that produces the LEAST amount of milk on that day and eat them. If there are more than one cow with minimal milk, John will be puzzled and will not eat any of them (Yeah! That's GREAT!!).

The i-th cow has a cycle of production Ti. That means, if it produces L unit milk on one day, it will also produce L unit after Ti days -- If it will not be eaten during these day :-). Though John is not a clever man, he doubts whether the cows will be eventually eaten up, so he asks for your help. Don't forget that he will offer you some nice beef for that!

Input

The first line of the input contains a single integer T, indicating the number of test cases. (1<=T<=50) Each test case begins with an integer N(1<=N<=1000), the number of cows. In the following N lines, each line contains an integer T_i(1<=T_i<=10), indicating the cycle of the i-th cow, then T_i integers M_j(0<=M_j<=250) follow, indicating the amount of milk it can produce on the j-th day.

Output

For each test case in the input, print a single line containing two integers C, D, indicating the number of cows that will NOT be eaten, and the number of days passed when the last cow is eaten. If no cow is eaten, the second number should be 0.

Sample Input

1
4
4 7 1 2 9
1 2
2 7 1
1 2

Sample Output

2 6
__________________________________________________________________________________________
Rujia Liu

目描述：

每牛的生牛奶的量按照期性生，每天生最少的牛掉，如果有相同少的牛，一也不。

最後剩下牛，而最後一是第天的。

目解法：

由於期最多 10 天，最小公倍 2520 天。

每次 2520 天，查找每一天是否有牛可以，直到某循任何一牛都不成。

#include <stdio.h>
// For the values: 10, 9, 8, 7, 6, 5, 4, 3, 2
// The LCM is: 2520
int main() {
    int testcase;
    int n, i, j, k;
    int T[1005][15], Tn[1005];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++) {
            scanf("%d", &Tn[i]);
            for(j = 0; j < Tn[i]; j++)
                scanf("%d", &T[i][j]);
        }
        int killed[1005] = {};
        int LCM = 2520, days = 0, lastDay = 0;
        int hasKilled = 1, sum;
        while(hasKilled) {
            hasKilled = 0;
          &nbp; for(i = 0; i < LCM; i++) {
                days++;
                int mn = 0xfffffff, mncnt = 0, mnidx;
                for(j = 0; j < n; j++) {
                    if(killed[j])    continue;
                    int v = T[j][i%Tn[j]];
                    if(v < mn)
                        mn = v, mncnt = 0, mnidx = j;
                    if(v == mn)
                        mncnt++;
                }
                if(mncnt == 1) {
                    killed[mnidx] = 1;
                    lastDay = days;
                    hasKilled = 1;
                }
            }
        }
        for(i = 0, sum = 0; i < n; i++)
            sum += killed[i] == 0;
        printf("%d %d\n", sum, lastDay);
    }
    return 0;
}