Zoj2235 poj1928 hdu1355 the peanuts

There is no greedy difficulty. The policy question has been given. Data is pre-processed and sorted from large to small by the number of peanuts.

/******************************************************************************* # Author : Neo Fung # Email : neosfung@gmail.com # Last modified: 2012-02-24 19:13 # Filename: ZOJ2235 POJ1928 HDU1355 The Peanuts.cpp # Description :  ******************************************************************************/#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX 3000using namespace std;struct NODE{  int x,y,p;  bool operator<(const NODE &t) const  {    return p>t.p;  }}node[MAX];int main(void){#ifdef DEBUG    freopen("../stdin.txt","r",stdin);  freopen("../stdout.txt","w",stdout); #endif  int ncases,n,m,time,temp;scanf("%d",&ncases);while(ncases--){scanf("%d%d%d",&n,&m,&time);int cnt=0;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){scanf("%d",&temp);if(temp){node[cnt].x=i;node[cnt].y=j;node[cnt++].p=temp;}}sort(node,node+cnt);int ans=0;temp=0;if(node[0].x*2+1<=time){ans=node[0].p;temp=node[0].x+1;for(int i=1;i<cnt;++i)if(temp+(abs(node[i].x-node[i-1].x)+abs(node[i].y-node[i-1].y)+1) + node[i].x<=time){temp+=(abs(node[i].x-node[i-1].x)+abs(node[i].y-node[i-1].y)+1);ans+=node[i].p;}elsebreak;}printf("%d\n",ans);}  return 0;}