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 501

using 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
="<<sx
<<endl;

//cout<<"a
[0] = "<<a
[0]<<endl;

//cout<<f[0][0][0]<<endl;

for (int x = 0; x <= sx
; x++)

for (int y = 0; y <= a
[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;

}

}