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February 14, 2018 19:29
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Study dynamic programming - back pack problem - Feb. 14, 2018
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/* | |
01背包问题 | |
背包九讲:http://www.cnblogs.com/jbelial/articles/2116074.html | |
其状态转移方程便是:f[i][v] = max{f[i-1][v], f[i-1][v-c[i]] + w[i]}。 | |
这个方程非常重要,基本上所有跟背包相关的问题的方程都是由它衍生出来的。 | |
**详细解释** | |
“将前 i 件物品放入容量为 v 的背包中” 这个子问题,若只考虑第 i 件物品的策略(放或不放),那么就 | |
可以转化为一个只牵扯前 i-1 件物品的问题。如果不放第 i 件物品,那么问题就转化为“前 i-1 件物品放入 | |
容量为 v 的背包中”;如果放第 i 件物品,那么问题就转化为“前 i - 1 件物品放入剩下的容量为 v - c[i]的背包中”, | |
此时能获得的最大价值就是 f [i - 1][v - c[i]]再加上通过放入第 i 件物品获得的价值 w[i]。 | |
*/ | |
int main() | |
{ | |
//int m = 120; | |
//int n = 5; | |
//vector<int> w = { 0, 40, 50, 70, 40, 20 }; | |
//vector<int> v = { 0, 10, 25, 40, 20, 10 }; | |
int m, n; //m重量,n数量 | |
while (cin >> m >> n) | |
{ | |
vector<int> w(n + 1, 0); | |
vector<int> v(n + 1, 0); | |
for (int i = 1; i <= n; i++) | |
{ | |
int tmp; | |
cin >> tmp; | |
w[i] = tmp; | |
} | |
for (int i = 1; i <= n; i++) | |
{ | |
int tmp; | |
cin >> tmp; | |
v[i] = tmp; | |
} | |
vector< vector<int> > vec(n + 1, vector<int>(m + 1, 0)); | |
for (int i = 1; i <= n; i++) | |
{ | |
for (int j = 1; j <= m; j++) | |
{ | |
if (w[i] > j) | |
vec[i][j] = vec[i - 1][j]; | |
else | |
{ | |
int tmp1 = v[i] + vec[i - 1][j - w[i]]; | |
int tmp2 = vec[i - 1][j]; | |
vec[i][j] = tmp1 > tmp2 ? tmp1 : tmp2; | |
} | |
} | |
} | |
double val = vec[n][m] * 0.1; | |
cout << val << endl; | |
} | |
system("pause"); | |
} |
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