ZOJ 1013 Great Equipment ---- DP

/***Zoj 1013 Great Equipment 主要思路   遍历所有可能的携带方案组合（x, y, z)其中， x 为携带的第0中设备数量，   y为携带的第1种设备数量， z为携带的第2种设备数量。      使用Dynamic progromming 来计算可能的携带方案 ****/#include <cstdlib>#include <iostream>#include <cstring>#define MAXN 101#define MAXQUANTITY 501using namespace std;inline int maxSets(int x, int y, int z, int q[3]){       x = x/q[0]; y = y/q[1]; z = z/q[2];       return x > y ? (y > z ? z : y) : (x > z ? z : x);}//前 i个车队带了x个第0种设备和 y个第1种设备后， 还能带的第2种设备数量 int f[2][MAXQUANTITY][MAXQUANTITY]; int main(int argc, char *argv[]){    int n;    int w[3], s[3], d[3], q[3];    //商队能带的最大重量， 尺寸     int wt[MAXN], sz[MAXN];            //前i个车队带了x个第0种设备后 ，   还能带的第1种设备数量     int a[MAXN][MAXQUANTITY];            int sx[MAXN];    int value;    int cases=1;    while (true)    {          cin>>n;          if (n == 0) break;          for (int i = 0; i < 3; i++)          cin>>w[i]>>s[i]>>d[i];          cin>>q[0]>>q[1]>>q[2];          cin>>value;                    /*计算前i个车队能带的第0种设备数量 */          sx[0] = 0;          for (int i = 1; i <= n; i++)          {              cin>>wt[i]>>sz[i];              int x1, x2;              sx[i] =  sx[i-1] +                 (( x1 = wt[i]/w[0]) > (x2 = sz[i] / s[0]) ? x2 : x1);          }          //for (int i = 0; i <= n; i++                    /*计算前 i个车队在带了x种第0种设备后， 还能带的第一种设备数量 */          memset(a[0], 0, sizeof(a[0]));          for (int i = 1; i <= n; i++)          {              for (int x = 0; x <= sx[i]; x++)              {                  a[i][x] = 0;                  //y: 第 i个车队带的第0种设备数量                   int y = 0;                  if (x > sx[i-1]) y = x - sx[i-1];                  for (; y <= sx[i] - sx[i-1]; y++)                  {                      int x1, x2;                      int num1 = ( x1 = (wt[i] - y*w[0])/w[1] )  > ( x2 = (sz[i] - y*s[0])/s[1] )                                    ? x2 : x1;                      if ( a[i-1][x-y] + num1 > a[i][x])                        a[i][x] = num1 + a[i-1][x-y];                  }              }          }          //           for (int i = 0; i <= n; i++)//          {//              for (int x = 0; x <= sx[i]; x++)//                cout<<a[i][x]<<" ";//              cout<<endl;//          }                    /*计算前i个车队在带了x种0设备， y种1设备后， 能带的2设备数量*/         // for (int x = 0; x < MAXQUANTITY; x++)          //  for (int y = 0; y < MAXQUANTITY; y++)          f[0][0][0] = 0;          for (int i = 1; i <= n; i++)          {              for (int x = 0; x <= sx[i]; x++)              {                  for (int y = 0; y <= a[i][x]; y++)                  {                      f[1][x][y] = 0;                      int t0, t1;                      int mx0 = ( t0 = wt[i]/w[0] ) > ( t1 = sz[i]/s[0] ) ? t1 : t0;                      //x0 第i个车队带的0设备数量， x1第i个车队带的1设备数量                       int x0 = 0;                      //当x总数超过前i-1个车队能带的0设备能力时， 第i车队应当至少带：                       if ( x > sx[i-1] ) x0 = x - sx[i-1];                                            for (; x0 <= mx0 && x0 <= x; x0++)                      {                          int mx1 = ( t0 = (wt[i] - x0*w[0])/w[1] ) > ( t1 = (sz[i] - x0*s[0])/s[1] )                                     ? t1 : t0;                          int x1 = 0;                          if ( y > a[i-1][x - x0] ) x1 = y - a[i-1][x - x0];                                     for (; x1 <= mx1 && x1 <= y; x1++)                          {                              int mx2 = ( t0 = (wt[i] - x0*w[0] - x1*w[1])/w[2] )                               > ( t1 = (sz[i] - x0*s[0] - x1*s[1])/s[2] ) ? t1 : t0;                                                              if (f[1][x][y] < f[0][x-x0][y-x1] + mx2)                                   f[1][x][y] = f[0][x-x0][y-x1] + mx2;                              // cout<<"x-x0 = "<<x - x0<<endl;                              // cout<<"y-x1  ="<<y - x1<<endl;                              // cout<<"f[1][x][y] = "<<f[1][x][y]<<endl;                               // cout<<"f[0][x][y] = "<<f[0][x-x0][y-x1]<<endl;                            }                      }                  }              }              for (int x = 0; x <= sx[i]; x++)                 for (int y = 0; y <= a[i][x]; y++)                    f[0][x][y] = f[1][x][y];           }           //cout<<"Here!/n";           //枚举遍历， 求最大值           int max = 0, mm, kk;           int xx, yy, zz;           //bool flag = value > q[0]*d[0]+q[1]*d[1]+q[1]*d[2];           //cout<<"sx[n]="<<sx[n]<<endl;           //cout<<"a[n][0] = "<<a[n][0]<<endl;           //cout<<f[0][0][0]<<endl;           for (int x = 0; x <= sx[n]; x++)              for (int y = 0; y <= a[n][x]; y++)                 for (int z = 0; z <= f[0][x][y]; z++)                 {                               kk = maxSets(x, y, z, q);                               xx = x - kk*q[0]; yy = y - kk*q[1];                               zz = z - kk*q[2];                               mm = xx*d[0] + yy*d[1] + zz*d[2] + kk*value;                               if ( mm > max ) max = mm;                 }           if (cases > 1) cout<<endl;           cout<<"Case "<<cases++<<": "<<max<<endl;             }}