題干：

In Zhejiang University Programming Contest, a team is called "couple team" if it consists of only two students loving each other. In the contest, the team will get a lovely balloon with unique color for each problem they solved. Since the girl would prefer pink balloon rather than black balloon, each color is assigned a value to measure its attractiveness. Usually, the boy is good at programming while the girl is charming. The boy wishes to solve problems as many as possible. However, the girl cares more about the lovely balloons. Of course, the boy's primary goal is to make the girl happy rather than win a prize in the contest.

Suppose for each problem, the boy already knows how much time he needs to solve it. Please help him make a plan to solve these problems in strategic order so that he can maximize the total attractiveness value of balloons they get before the contest ends. Under this condition, he wants to solve problems as many as possible. If there are many ways to achieve this goal, he needs to minimize the total penalty time. The penalty time of a problem is equal to the submission time of the correct solution. We assume that the boy is so clever that he always submit the correct solution.

Input

The first line of input is an integer?N?(N?< 50) indicating the number of test cases. For each case, first there is a line containing 2 integers?T?(T?<= 1000) and?n?(n?<= 50) indicating the contest length and the number of problems. The next line contains?n?integers and the?i-th integer?ti?(ti?<= 1000) represents the time needed to solve the ith problem. Finally, there is another line containing?n?integers and the?i-th integer?vi?(vi?<= 1000) represents the attractiveness value of the?i-th problem. Time is measured in minutes.

Output

For each case, output a single line containing 3 integers in this order: the total attractiveness value, the number of problems solved, the total penalty time. The 3 integers should be separated by a space.

Sample Input

2 300 10 10 10 10 10 10 10 10 10 10 10 1 2 3 4 5 6 7 8 9 10 300 10 301 301 301 301 301 301 301 301 301 301 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000

Sample Output

55 10 550 0 0 0

題目大意：

給出比賽時間和題目數，并給出每道題目的耗時以及獲得的價值，找出所能獲得的最大價值，獲得這個最大價值下的最多題目數以及最少罰時，這里的罰時理解的半天，舉第一個例子：A掉第一道題需要用時10分鐘，然后第二道題也需要10分鐘，即第二道題得在20分鐘的時候A掉，所以兩道題的罰時是10+20。

解題報告：

? 直接用他給定的優先級就行了，按照優先級排序然后按照這個優先級順序進行重載小于號然后做背包就行了。好題一只。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; int n,T; struct Node {ll val,t; } node[MAX]; struct N {ll val,num,t;N(){}N(ll val,ll num,ll t):val(val),num(num),t(t){}bool operator<(const N b)const{if(val != b.val) return val < b.val;if(num != b.num) return num < b.num;return t > b.t;} } dp[MAX]; bool cmp(Node a,Node b) {return a.t < b.t; } int main() {int t;cin>>t;while(t--) {scanf("%d%d",&T,&n);memset(dp,0,sizeof dp);for(int i = 1; i<=n; i++) scanf("%lld",&node[i].t);for(int i = 1; i<=n; i++) scanf("%lld",&node[i].val);sort(node+1,node+n+1,cmp);N ans = N(0,0,0);for(int i = 1; i<=n; i++) {for(int j = T; j>=node[i].t; j--) {N pre = dp[j-node[i].t];N tmp = N(pre.val + node[i].val,pre.num + 1,pre.t + j);if(dp[j] < tmp) dp[j] = tmp;if(ans < dp[j]) ans = dp[j];}}printf("%lld %lld %lld\n",ans.val,ans.num,ans.t);}return 0 ; }

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