编程算法 - 完全背包问题 代码(C)

完全背包问题 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 有n个重量和价值分别为w,v的物品, 从这些物品中挑选出总重量不超过W的物品, 求所有挑选方案中价值总和的最大值. 
*每件物品可以挑选任意多件.

动态规划: 每次选取最大的组合, 加入到数组, 第一种时间复杂度O(nW^2), 第二种时间复杂度O(nW).
解法1, max()部分表明, 要么来源于上面, 要么来源于前面.

代码:
/*
* main.cpp
*
* Created on: 2014.7.17
* Author: spike
*/

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <memory.h>
#include <limits.h>

#include <utility>
#include <queue>
#include <algorithm>

using namespace std;

class Program {
static const int MAX_N = 100;

int n=4, W=5;
int w[MAX_N] = {2,1,3,2}, v[MAX_N]={3,2,4,2};
int dp[MAX_N+1][MAX_N+1];
public:
void solve() {
for (int i=0; i<n; ++i) {
for (int j=0; j<=W; j++) {
for (int k=0; k*w[i] <= j; k++) {
dp[i+1][j] = max(dp[i+1][j], dp[i][j-k*w[i]]+k*v[i]);
}
}
}
printf("result = %d\n", dp
[W]);
}

void solve1() {
for (int i=0; i<n; ++i) {
for (int j=0; j<=W; j++) {
if (j<w[i]) {
dp[i+1][j] = dp[i][j];
} else {
dp[i+1][j] = max(dp[i][j], dp[i+1][j-w[i]]+v[i]);
}
}
}
printf("result = %d\n", dp
[W]);
}
};

int main(void)
{
Program P;
P.solve1();
return 0;
}


输出:
result = 10


为了节约内存, 可以使用一维数组进行求解.

代码:
/*
* main.cpp
*
* Created on: 2014.7.17
* Author: spike
*/

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <memory.h>
#include <limits.h>

#include <utility>
#include <queue>
#include <algorithm>

using namespace std;

class Program {
static const int MAX_N = 100;

int n=3, W=7;
int w[MAX_N] = {3,4,2}, v[MAX_N]={4,5,3};
int dp[MAX_N+1];
public:
void solve() {
memset(dp, 0, sizeof(dp));
for (int i=0; i<n; ++i) {
for (int j=w[i]; j<=W; j++) {
dp[j] = max(dp[j], dp[j-w[i]]+v[i]);
}
}
printf("result = %d\n", dp[W]);
}
};

int main(void)
{
Program P;
P.solve();
return 0;
}


输出:
result = 10


可以讲两个数组滚动使用, 节省内存.
代码:
/*
* main.cpp
*
* Created on: 2014.7.17
* Author: spike
*/

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <memory.h>
#include <limits.h>

#include <utility>
#include <queue>
#include <algorithm>

using namespace std;

class Program {
static const int MAX_N = 100;

int n=3, W=7;
int w[MAX_N] = {3,4,2}, v[MAX_N]={4,5,3};
int dp[2][MAX_N+1];
public:
void solve() {
memset(dp,0,sizeof(dp));
for (int i=0; i<n; ++i) {
for (int j=0; j<=W; j++) {
if (j<w[i]) {
dp[(i+1)&1][j] = dp[i&1][j];
} else {
dp[(i+1)&1][j] = max(dp[i&1][j], dp[(i+1)&1][j-w[i]]+v[i]);
}
}
}
printf("result = %d\n", dp[n&1][W]);
}
};

int main(void)
{
Program P;
P.solve();
return 0;
}


输出:
result = 10


当重量可选范围很大, 价值可选范围小的时候, 可以使根据价值进行动态规划.

代码:
/*
* main.cpp
*
* Created on: 2014.7.17
* Author: spike
*/

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <memory.h>
#include <limits.h>

#include <utility>
#include <queue>
#include <algorithm>

using namespace std;

class Program {
static const int MAX_N = 100;
static const int MAX_V = 4;
const int INF = INT_MAX>>2;

int n=4, W=5;
int w[MAX_N] = {2,1,3,2}, v[MAX_N] = {3,2,4,2};

int dp[MAX_N+1][MAX_N*MAX_V+1];
public:
void solve() {
fill(dp[0], dp[0]+MAX_N*MAX_V+1, INF);

dp[0][0] = 0;
for (int i=0; i<n; ++i) {
for (int j=0; j<=MAX_N*MAX_V; j++) {
if (j<v[i]) {
dp[i+1][j] = dp[i][j];
} else {
dp[i+1][j] = min(dp[i][j], dp[i][j-v[i]]+w[i]);
}
}
}
int res = 0;
for (int i=0; i<=MAX_N*MAX_V; ++i) if (dp
[i]<=W) res = i;
printf("result = %d\n", res);
}
};

int main(void)
{
Program P;
P.solve();
return 0;
}


输出:
result = 7