您的位置:首页 > 其它

动态规划之01背包,完全背包,多重背包模板

2017-08-24 15:46 483 查看
自己整理的模板,记录一下:

#include <iostream>
#include <vector>
#include <deque>
#include <algorithm>

using namespace std;

const int M = 500;
const int N = 500;

#pragma region"01背包问题"
// 未优化版
int pack01(vector<int>& weight, vector<int>& value, int sum, int n)
{
int dp[M]
;
for (int i = 0; i <= sum; ++i)
{
dp[0][i] = i >= weight[0] ? value[0] : 0;
}
for (int i = 1; i < n; ++i)
{
for (int v = weight[i]; v <= sum; ++v)
{
dp[i][v] = max(dp[i - 1][v], dp[i - 1][v - weight[i]] + value[i]);
}
}
return dp[n - 1][sum];
}

// 空间优化版(滚动数组)
int pack01optimize(vector<int>& weight, vector<int>& value, int sum, int n)
{
int dp
;
for (int i = 0; i <= sum; ++i)
{
dp[i] = i >= weight[0] ? value[0] : 0;
}
for (int i = 1; i < n; ++i)
{
for (int v = sum; v >= weight[i]; --v)
{
dp[v] = max(dp[v], dp[v - weight[i]] + value[i]);
}
}
return dp[sum];
}

#pragma endregion

#pragma region"完全背包问题"
// 完全背包问题(每个物品可以选无数次)
int packComplete(vector<int>& weight, vector<int>& value, int sum, int n)
{
int dp
;
for (int i = 0; i <= sum; ++i)
{
dp[i] = i >= weight[0] ? (dp[i - weight[0]] + value[0]) : 0;
}
for (int i = 1; i < n; ++i)
{
for (int v = weight[i]; v <= sum; ++v) // 和01背包的区别
{
dp[v] = max(dp[v], dp[v - weight[i]] + value[i]);
}
}
return dp[sum];
}
#pragma endregion

#pragma region"多重背包"
#pragma region"二进制优化版"
void oneZeroPack(int dp[], int sum, int weight, int value)
{
for (int i = sum; i >= weight; --i)
{
dp[i] = max(dp[i], dp[i - weight] + value);
}
}

void completePack(int dp[], int sum, int weight, int value)
{
for (int i = weight; i <= sum; ++i)
{
dp[i] = max(dp[i], dp[i - weight] + value);
}
}
// 把每个物体分成1, 2, 4 ..., 2^(k - 1), cnt - 2^k + 1个,然后转变成01背包问题,其中k为满足cnt - 2^k + 1 > 0的最大整数
int MultiPack(vector<int>& weight, vector<int>& value, vector<int>& cnt, int sum, int n)
{
int dp
= {};
for (int i = 0; i < n; ++i)
{
if (weight[i] * cnt[i] >= sum)
{
completePack(dp, sum, weight[i], value[i]);
}
else
{
int k = 1;
for (k = 1; k <= cnt[i]; k <<= 1)
{
oneZeroPack(dp, sum, weight[i], value[i]);
cnt[i] -= k;
}
if (cnt[i])
{
oneZeroPack(dp, sum, weight[i], value[i]);
}
}
}
return dp[sum];
}
#pragma endregion

#pragma region"单调队列版"
struct Pack
{
int num, value;
Pack(int _num, int _value) : num(_num), value(_value)
{
}
};
deque<Pack> q;
int multiPackQueue(vector<int>& weight, vector<int>& value, vector<int>& cnt, int sum, int n)
{
int dp
;
for (int i = 0; i < n; ++i)
{
if (cnt[i] > (sum / weight[i])) cnt[i] = sum / weight[i];
for (int d = 0; d < weight[i]; ++d)
{
q.clear();
for (int j = 0; j <= (sum - d) / weight[i]; ++j)
{
int tValue = dp[d + j * weight[i]] - j * value[i];
while (!q.empty() && q.back().value <= tValue) q.pop_back();
q.push_back(Pack(j, tValue));
while (!q.empty() && q.front().num < j - cnt[i]) q.pop_front();
dp[d + j * weight[i]] = q.front().value + j * value[i];
}
}
}
return dp[sum];
}

#pragma endregion
#pragma endregion
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  动态规划