The Peanuts
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 6500 | Accepted: 2736 |
Description
Mr. Robinson and his pet monkey Dodo love peanuts very much. One day while they were having a walk on a country road, Dodo found a sign by the road, pasted with a small piece of paper, saying "Free Peanuts Here! " You can imagine how happy Mr. Robinson and Dodo were.
There was a peanut field on one side of the road. The peanuts were planted on the intersecting points of a grid as shown in Figure-1. At each point, there are either zero or more peanuts. For example, in Figure-2, only four points have more than zero peanuts, and the numbers are 15, 13, 9 and 7 respectively. One could only walk from an intersection point to one of the four adjacent points, taking one unit of time. It also takes one unit of time to do one of the following: to walk from the road to the field, to walk from the field to the road, or pick peanuts on a point.
According to Mr. Robinson's requirement, Dodo should go to the plant with the most peanuts first. After picking them, he should then go to the next plant with the most peanuts, and so on. Mr. Robinson was not so patient as to wait for Dodo to pick all the peanuts and he asked Dodo to return to the road in a certain period of time. For example, Dodo could pick 37 peanuts within 21 units of time in the situation given in Figure-2.
Your task is, given the distribution of the peanuts and a certain period of time, tell how many peanuts Dodo could pick. You can assume that each point contains a different amount of peanuts, except 0, which may appear more than once.
There was a peanut field on one side of the road. The peanuts were planted on the intersecting points of a grid as shown in Figure-1. At each point, there are either zero or more peanuts. For example, in Figure-2, only four points have more than zero peanuts, and the numbers are 15, 13, 9 and 7 respectively. One could only walk from an intersection point to one of the four adjacent points, taking one unit of time. It also takes one unit of time to do one of the following: to walk from the road to the field, to walk from the field to the road, or pick peanuts on a point.
According to Mr. Robinson's requirement, Dodo should go to the plant with the most peanuts first. After picking them, he should then go to the next plant with the most peanuts, and so on. Mr. Robinson was not so patient as to wait for Dodo to pick all the peanuts and he asked Dodo to return to the road in a certain period of time. For example, Dodo could pick 37 peanuts within 21 units of time in the situation given in Figure-2.
Your task is, given the distribution of the peanuts and a certain period of time, tell how many peanuts Dodo could pick. You can assume that each point contains a different amount of peanuts, except 0, which may appear more than once.
Input
The first line of input contains the test case number T (1 <= T <= 20). For each test case, the first line contains three integers, M, N and K (1 <= M, N <= 50, 0 <= K <= 20000). Each of the following M lines contain N integers. None of the integers will exceed 3000. (M * N) describes the peanut field. The j-th integer X in the i-th line means there are X peanuts on the point (i, j). K means Dodo must return to the road in K units of time.
Output
For each test case, print one line containing the amount of peanuts Dodo can pick.
Sample Input
2 6 7 21 0 0 0 0 0 0 0 0 0 0 0 13 0 0 0 0 0 0 0 0 7 0 15 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 6 7 20 0 0 0 0 0 0 0 0 0 0 0 13 0 0 0 0 0 0 0 0 7 0 15 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0
Sample Output
37 28
1 /* 2 功能Function Description: POJ 1928 The Peanuts 3 开发环境Environment: DEV C++ 4.9.9.1 4 技术特点Technique: 5 版本Version: 6 作者Author: 可笑痴狂 7 日期Date: 20120727 8 备注Notes: 结构体+排序 9 */ 10 /* 11 //代码一:冒泡排序 196K 141MS 12 #include<stdio.h> 13 #include<math.h> 14 15 struct point 16 { 17 int x; 18 int y; 19 int num; 20 }node[2510],temp; 21 22 int init(int m,int n) //初始化存储有花生的节点,并返回节点个数 23 { 24 int t,i,j,k=0; 25 for(i=1;i<=m;++i) 26 for(j=1;j<=n;++j) 27 { 28 scanf("%d",&t); 29 if(t) 30 { 31 node[k].x=j; 32 node[k].y=i; 33 node[k].num=t; 34 ++k; 35 } 36 } 37 return k; 38 } 39 40 int main() 41 { 42 int T,m,n,i,j,time,k; 43 long sum; 44 scanf("%d",&T); 45 while(T--) 46 { 47 scanf("%d%d%d",&m,&n,&time); 48 sum=0; 49 k=init(m,n); 50 for(i=k-1;i>0;--i) 51 for(j=0;j<i;++j) 52 { 53 if(node[j].num<node[j+1].num) 54 { 55 temp=node[j]; 56 node[j]=node[j+1]; 57 node[j+1]=temp; 58 } 59 } 60 time-=node[0].y; 61 for(i=0;i<k;++i) 62 { 63 if(time>=node[i].y+1) 64 { 65 sum+=node[i].num; 66 time-=abs(node[i].x-node[i+1].x)+abs(node[i].y-node[i+1].y)+1; 67 } 68 else 69 break; 70 } 71 printf("%d\n",sum); 72 } 73 return 0; 74 } 75 76 */ 77 78 //代码二:用qsort实现快排 196K 16MS 79 #include<stdio.h> 80 #include<math.h> 81 #include<stdlib.h> 82 83 struct point 84 { 85 int x; 86 int y; 87 int num; 88 }node[2510],temp; 89 90 int init(int m,int n) //初始化存储有花生的节点,并返回节点个数 91 { 92 int t,i,j,k=0; 93 for(i=1;i<=m;++i) 94 for(j=1;j<=n;++j) 95 { 96 scanf("%d",&t); 97 if(t) 98 { 99 node[k].x=j; 100 node[k].y=i; 101 node[k].num=t; 102 ++k; 103 } 104 } 105 return k; 106 } 107 108 int comp(const void *p1,const void *p2) 109 { 110 return (*(struct point *)p2).num-(*(struct point *)p1).num; 111 } 112 113 int main() 114 { 115 int T,m,n,i,time,k; 116 long sum; 117 scanf("%d",&T); 118 while(T--) 119 { 120 scanf("%d%d%d",&m,&n,&time); 121 sum=0; 122 k=init(m,n); 123 qsort(node,k,sizeof(struct point),comp); 124 time-=node[0].y; 125 for(i=0;i<k;++i) 126 { 127 if(time>=node[i].y+1) 128 { 129 sum+=node[i].num; 130 time-=abs(node[i].x-node[i+1].x)+abs(node[i].y-node[i+1].y)+1; 131 } 132 else 133 break; 134 } 135 printf("%d\n",sum); 136 } 137 return 0; 138 }