poj 1042 钓鱼问题 枚举+贪心

#include <stdio.h>
#include <string.h>
int f[25];          //第一个五分钟能钓到的鱼 
int curF[25];       //目前各湖能钓到的鱼数 
int d[25];          //每五分钟鱼量减少的数目 
int ans[25];        //能钓到最大鱼数时各湖停留的时间 
int ansTmp[25];     
int t[25];          //从i走到i+1所需要的时间        
int n;
int times;          //总的钓鱼时间 
int resTimes;       //总的钓鱼时间 - 行走于各湖之间的时间 
int fish;           //能钓的最大鱼数 
int fishTmp;
int main() {
    while(scanf("%d", &n) != EOF && n != 0) {
          scanf("%d", ×);
          for(int i = 0; i < n; i++)
              scanf("%d", &f[i]);
          for(int i = 0; i < n; i++)
              scanf("%d", &d[i]);
          for(int i = 1; i < n; i++)
              scanf("%d", &t[i]);
          t[0] = 0;
          times *= 12;
          resTimes = 0;
          fish = 0;
          memset(ans, 0, sizeof(ans));
          for(int len = 1; len <= n; len++) {
              memcpy(curF, f, sizeof(f));
              memset(ansTmp, 0, sizeof(ansTmp));
              resTimes += t[len - 1];
              fishTmp = 0;
              for(int i = 1; i <= times - resTimes; i++) {
                  int _maxi = 0;
                  for(int j = 1; j < len; j++)       
                      if(curF[j] > curF[_maxi])
                          _maxi = j;
                  ansTmp[_maxi]++;
                  fishTmp += curF[_maxi];
                  curF[_maxi] -= d[_maxi];
                  if(curF[_maxi] < 0)
                      curF[_maxi] = 0;
              }
              if(fishTmp > fish) {
                  fish = fishTmp;
                  memcpy(ans, ansTmp, sizeof(ansTmp));
              }else if(fishTmp == fish) {
                for(int i = 0; i < len; i++) {
                    if(ans[i] > ansTmp[i]) break;        //之前忘了这句，导致多解的时候可能出错，WA3次后发现 
                    if(ans[i] < ansTmp[i]) {
                        memcpy(ans, ansTmp, sizeof(ansTmp));
                        break;      
                    }        
                }      
              }
          }
          printf("%d", ans[0] * 5);
          for(int i = 1; i < n; i++)
             printf(", %d", ans[i] * 5);
          printf("/nNumber of fish expected: %d/n/n", fish);
    }
    return 0;
}