Nyoj 747 ant financial problems (III)

Ant financial problems (III)Time Limit: 2000 MS | memory limit: 65535 kb difficulty: 4

Description

Ant finally moved as many ingredients as possible back home, and now started the chef program.

Known to allNFood Ingredients, each of which has a delicious tasteAIAnd freshnessBiIf ant financialTAlways changeIThe sample food is successfully cooked.Ai-T * biOf course, it takes time to cook with the I-th foodCi.

As we all know, ant's cooking is not very good, so he needs you to design a cooking solution so that TComplete the most delicious index.

 

Input

There are multiple groups of test data.
The first line is two positive integers, indicating the ant's cooking time t and the number of ingredients n. (N <= 50, 1 <= T <= 100000 ).
Next n rows, each row has three numbers: AI, Bi, and CI. It represents the taste, freshness, and time spent cooking with the food. (0 <AI, Bi, CI <= 100000 ).

Output

Output A number indicating the maximum delicious Index

Sample Input

6 1200 5 1

Sample output

195

Source

Ant Series

Uploaded

ACM _ Li rubing

Problem solving: This question is the same as the physical examination in the queue .... ^_^ .....

Assume that the first food item is preferred.

A1-c1b1 + A2-(C1 + C2) B2> a2-c2b2 + A1-(C1 + C2) B 1

=> c1b2 < c2b1

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 struct node {12     int a,b,c;13 } p[52];14 int dp[100010];15 bool cmp(const node &x,const node &y) {16     return y.c*x.b > x.c*y.b;17 }18 int main() {19     int t,n,i,j,ans;20     while(~scanf("%d %d",&t,&n)) {21         for(i = 0; i < n; i++)22             scanf("%d %d %d",&p[i].a,&p[i].b,&p[i].c);23         sort(p,p+n,cmp);24         memset(dp,0,sizeof(dp));25         for(ans = i = 0; i < n; i++) {26             for(j = min(t,p[i].a/p[i].b); j >= p[i].c; j--) {27                 dp[j] = max(dp[j],dp[j-p[i].c]+p[i].a-j*p[i].b);28                 if(dp[j] > ans) ans = dp[j];29             }30         }31         printf("%d\n",ans);32     }33     return 0;34 }

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