AcWing 2. 01背包问题
原题链接
简单
C++ 代码
#二维数组原始版
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1010;
int w[N] = { 0 };//权重,即价值
int v[N] = { 0 };//体积
int f[N][N] = { {0},{0} };//状态表达式
int main()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++i) {
cin >> v[i] >> w[i];
}
for (int i = 1; i <= n; ++i) {//f[i,j]为在i件物品中选体积不大于j的物品的最大总价值
for (int j = 1; j <= m; ++j) {
if (v[i] > j)//如果第I个物品体积就大于背包体积,则该物品被放弃
f[i][j] = f[i - 1][j];
else {
f[i][j] = max(f[i - 1][j] , f[i - 1][j - v[i]] + w[i]);//状态转移方程,第i个物品只有选与不选两种操作,在其中选取使得结果最大的情况
}
}
}
printf("%d\n", f[n][m]);
return 0;
}
#一维数组优化版(神奇的滚动数组)
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1010;
int w[N] = { 0 };
int v[N] = { 0 };
int f[N] = { 0 };
int main()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++i) {
cin >> v[i] >> w[i];
}
for (int i = 1; i <= n; ++i) {
for (int j = m; j >= v[i]; --j) {
f[j] = max(f[j], f[j - v[i]] + w[i]);
}
}
printf("%d\n", f[m]);
return 0;
}