标签：cannot capacity USA Contest int Regional objects knapsack remaining
题目描述
We have a knapsack of integral capacity and some objects of assorted integral sizes. We attempt to fill the knapsack up, but unfortunately, we are really bad at it, so we end up wasting a lot of space that can’t be further filled by any of the remaining objects. In fact, we are optimally bad at this!How bad can we possibly be?
figure out the least capacity we can use where we cannot place any of the remaining objects in the knapsack. For example, suppose we have 3 objects with weights 3, 5 and 3, and our knapsack has capacity 6. If we foolishly pack the object with weight 5 first, we cannot place either of the other two objects in the knapsack. That’s the worst we can do, so 5 is the answer.
输入
The first line of input contains two integers n (1 ≤ n ≤ 1,000) and c (1 ≤ c ≤ 105 ), where n is the number of objects we want to pack and c is the capacity of the knapsack.Each of the next n lines contains a single integer w (1 ≤ w ≤ c). These are the weights of the objects.
输出
Output a single integer, which is the least capacity we can use where we cannot place any of the remaining objects in the knapsack.样例输入 Copy
3 6 3 5 3
样例输出 Copy
5
这个题只有1-9九个数字，我们可以求一个9的全排列，但是我们需要check一个排列行不行
一种容易想到的方法就是O（n）扫边字符串就可以了
但是这样复杂度并不能通过
这里继续考虑这九种数字
每次移动都是在九种数字的任意两种之间
那么我们处理出相邻的两数字出现的次数
然后check的时候只需要考虑对于一个排列枚举两种数字，然后算出距离×出现次数的和就可以了
int a[10],ans = inf,p[maxn],pos[10];
int len,num[66][66] ;
char s[maxn];
int cal() {
int temp=0;
for(int i=1 ; i<=9 ; i++) pos[a[i]] = i;
for(int i=1 ; i<=9 ; i++) for(int j=1 ; j<=9 ; j++) temp+=abs(pos[i]-pos[j])*num[i][j];
return temp+len+abs(pos[p[1]] - pos[a[1]] );
}
int main() {
scanf("%s",s+1);
len = strlen(s+1);
rep(i,1,len ) p[i] = s[i]-'0';
for(int i=1 ; i<len ; i++)num[p[i]][p[i+1]]++;
rep(i,1,9) a[i] = i;
do {
ans = min(ans,cal());
} while(next_permutation(a+1,a+1+9));
out(ans);
return 0;
}
View Code
标签：cannot,capacity,USA,Contest,int,Regional,objects,knapsack,remaining 来源： https://www.cnblogs.com/UpMing/p/14645290.html
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