Time
Limit: 1000/1000 MS (Java/Others) Memory Limit:
32768/32768 K (Java/Others)
Total Submission(s):
14966 Accepted Submission(s):
4235
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 int cmp(const void *a ,const void *b) 5 { 6 return *(int *)b -*(int *)a; //从大到小 7 } 8 int main() 9 { 10 int n,tiny_num,i,j,k; 11 float q; 12 char str[12],tem; 13 float dp[35],val[35],num,tol_num[25]; 14 bool tag; 15 while(scanf("%f%d",&q,&n),n) 16 { 17 k=0; 18 memset(val,0,sizeof(val)); 19 while(n--) 20 { 21 scanf("%d",&tiny_num); 22 tag=false; 23 memset(tol_num,0,sizeof(tol_num)); 24 while(tiny_num--) 25 { 26 scanf("%s",str); 27 sscanf(str,"%c:%f",&tem,&num); 28 tol_num[tem-‘A‘]+=num; 29 if(tem!=‘A‘&&tem!=‘B‘&&tem!=‘C‘) 30 tag=true; 31 for(i=0;i<3;i++) 32 { 33 if(tol_num[i]>600.00) 34 tag=true; 35 } 36 } 37 float tol=tol_num[0]+tol_num[1]+tol_num[2]; 38 if(tol>1000.00) 39 tag=true ; 40 if(!tag) val[k++]=tol; 41 } 42 qsort(val,k,sizeof(val[0]),cmp); 43 memset(dp,0,sizeof(dp)); 44 for(i=0;i<k;i++) 45 { 46 for(j=k;j>0;j--) 47 { 48 if(dp[j-1]+val[i]<=q&&dp[j]<dp[j-1]+val[i]) 49 dp[j]=dp[j-1]+val[i]; 50 } 51 } 52 printf("%.2f\n",dp[k]); 53 } 54 return 0; 55 }
HDUOJ -----1864 最大报销额(动态规划),布布扣,bubuko.com
原文:http://www.cnblogs.com/gongxijun/p/3642933.html