用优先队列BFS一遍即可,
每个节点分别记录 当前难度,加上下一个以后的难度,和下一个为哪道题
队列优先弹出加上下一个以后难度最小的
#include "stdio.h"
#include "string.h"
#include "algorithm"
#include "queue"
using namespace std;
struct node
{
int now,next,id;
bool friend operator<(node n1,node n2)
{
return n2.next<n1.next;
}
};
int a[10010],n,m;
void bfs()
{
int i,cnt;
priority_queue<node>q;
node cur,next;
scanf("%d%d",&n,&m);
for (i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
cur.now=0;
cur.next=a[0];
cur.id=0;
q.push(cur);
cnt=0;
while (cnt<m)
{
cur=q.top();
q.pop();
if (cur.id>=n) continue;
next.now=cur.now;
next.next=cur.now+a[cur.id+1];
next.id=cur.id+1;
q.push(next);
next.now=cur.next;
next.next=cur.next+a[cur.id+1];
next.id=cur.id+1;
q.push(next);
cnt++;
}
for (i=0;!q.empty();i++)
{
a[i]=q.top().now;
q.pop();
}
sort(a,a+m);
printf("%d\n",a[m-1]);
}
int main()
{
int Case,i;
scanf("%d",&Case);
for (i=1;i<=Case;i++)
{
printf("Case #%d: ",i);
bfs();
}
return 0;
}