两种0-1背包的动态规划解决方法。下面第一种可以获得最优解，但是占空间，第二种节省空间。状态转移方程见函数
例题参见poj3624
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
/*------default---------*/
const int maxn = 4 * 1e3 + 7;
int v[maxn], w[maxn], dp[maxn][maxn];
//max(m(i + 1, j), m(i + 1, j - w[i]) + vi); j >= w[i] 当背包装得下w[i]，不装 or 装 ？
void Knapsack(int value[], int weght[], int c, int n, int m[][maxn]) {
//m[i][j]表示背包容量为j，可选物品为[i:n]时0-1背包的最优值
int jMax = min(weght[n] - 1, c);
for (int j = 0; j <= jMax; ++j)
m[n][j] = 0;
for (int j = weght[n]; j <= c; ++j)
m[n][j] = value[n];
for (int i = n - 1; i >= 1; --i) {
int jMax = min(weght[i] - 1, c);
for (int j = 0; j <= jMax; ++j)
m[i][j] = m[i + 1][j];
for (int j = weght[i]; j <= c; ++j)
m[i][j] = max(m[i + 1][j], m[i + 1][j - weght[i]] + value[i]);
}
}
void Traceback(int m[][maxn], int w[], int c, int n, int x[]) {
for (int i = 1; i < n; ++i) {
if (m[i][c] == m[i + 1][c])
x[i] = 0;
else {
x[i] = 1;
c -= w[i];
}
x[n] = m[n][c] ? 1 : 0;
}
}
int main()
{
int n, c;
while (~scanf("%d%d", &n, &c)) {
for (int i = 1; i <= n; ++i)
scanf("%d%d", w + i, v + i);
Knapsack(v, w, c, n, dp);
printf("%d\n", dp[1][c]);
}
return 0;
}
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void Knapsack(int w[], int v[], int n, int c, int dp[]) {
memset(dp, 0, sizeof dp);
for (int i = 1; i <= n; ++i)
for (int j = c; j >= w[i]; --j)
dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
}
//dp[j]表示背包容量为j时0-1背包的最优价值。数组范围为背包容量。
# by wjqin
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