标签：混合背包 style back float tor 循环 max 优化 来源
1.01背包
#include<iostream>
#include <cstring>
using namespace std;
const int N = 1010;
int f[N];
int main()
{
int n,m;
while(cin >> n >> m){
int v,w;
for (int i= 0 ; i < n;i++){
cin >> v >> w;
for(int j = m; j >= v;j--){
f[j] = max(f[j],f[j - v]+ w);
}
}
cout << f[m] << endl;
}
return 0;
}
2.完全背包
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int main()
{
const int N = 1010;
vector<int> dp(N);
vector<int> v(N);
vector<int> w(N);
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 = v[i]; j <= m; j++)
{
dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
}
cout << dp[m] << endl;
}
3.多重背包I
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int n, m;
const int N = 110;
int f[N];
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
int v, w, s;
cin >> v >> w >> s;
for (int j = m; j >= 0; j--) {
for (int k = 1; k <= s && k * v <= j; k++) {
f[j] = max(f[j], f[j - k * v] + k * w);
}
}
}
cout << f[m] << endl;
return 0;
}
4.多重背包II （二进制优化版本）
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int N = 2010;
int f[N];
struct Goods
{
int v;
int w;
};
int main() {
int n, m;
vector<Goods> goods;
cin >> n >> m;
for (int i = 0; i < n; i++) {
int v, w, s;
cin >> v >> w >> s;
for (int k = 1; k <= s; k *= 2) {
s -= k;
goods.push_back({ v * k,w * k });
}
if (s > 0) goods.push_back({ v * s,w * s }); // 如果 s还有剩余的
}
for (auto good : goods) {
for (int j = m; j >= good.v; j--) {
f[j] = max(f[j], f[j - good.v] + good.w); //转化为01背包
}
}
cout << f[m] << endl;
return 0;
}
5.混合背包
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int f[N];
struct Thing {
int kind;
int v, w;
};
vector<Thing> things;
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
int v, w, s;
cin >> v >> w >> s;
if (s < 0) { things.push_back({ -1,v,w }); } // 背包情况
else if (s == 0) { // 完全背包
things.push_back({ 0,v,w });
}
else { // 多重背包利用二进制优化转 01背包
for (int k = 1; k <= s; k *= 2) {
s -= k;
things.push_back({ -1,k * v,k * w });
}
if (s > 0) {
things.push_back({ -1,s * v,s * w });
}
}
}
for (auto thing : things) {
if (thing.kind < 0) { // 01 背包从大到小遍历
for (int j = m; j >= thing.v; j--) {
f[j] = max(f[j], f[j - thing.v] + thing.w);
}
}
else { // 完全背包，从小到大遍历
for (int j = thing.v; j <= m; j++) {
f[j] = max(f[j], f[j - thing.v] + thing.w);
}
}
}
cout << f[m] << endl;
return 0;
}
6.二维背包费用问题（两个约束——体积和重量约束）
#include <iostream>
#include<algorithm>
#include <cstring>
using namespace std;
const int N = 110;
int n, v, m;
int f[N][N];
int main() {
cin >> n >> v >> m;
for (int i = 0; i < n; i++) {
int a, b, c;
cin >> a >> b >> c;
for (int j = v; j >= a; j--) {
for (int k = m; k >= b; k--) {
f[j][k] = max(f[j][k], f[j - a][k - b] + c);
}
}
}
cout << f[v][m] << endl;
return 0;
}
7.分组背包
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 110;
int f[N], v[N], w[N];
int n, m;
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) { // 循环物品组数
int s;
cin >> s; // 当前组物品数量
for (int j = 0; j < s; j++) {
cin >> v[j] >> w[j]; // 当前组内物品数的体积和价值
}
for (int j = m; j >= 0; j--) {
for (int k = 0; k < s; k++) { // 同一组内的物品只能选一个
if (j >= v[k]) {
f[j] = max(f[j], f[j - v[k]] + w[k]);
}
}
}
}
cout << f[m] << endl;
return 0;
}
8.有依赖的背包问题
代码来源：https://www.acwing.com/solution/acwing/content/8316/
#include<iostream>
#include<vector>
using namespace std;
int f[110][110];//f[x][v]表达选择以x为子树的物品，在容量不超过v时所获得的最大价值
vector<int> g[110];
int v[110], w[110]; //物品的体积、价值和依赖的物品编号(父节点）
int n, m, root;
int dfs(int x)
{
for (int i = v[x]; i <= m; i++) f[x][i] = w[x];//点x必须选，所以初始化f[x][v[x] ~ m]= w[x]
for (int i = 0; i < g[x].size(); i++)
{
int y = g[x][i];
dfs(y);
for (int j = m; j >= v[x]; j--)//j的范围为v[x]~m, 小于v[x]无法选择以x为子树的物品
{
for (int k = 0; k <= j - v[x]; k++)//分给子树y的空间不能大于j-v[x],不然都无法选根物品x
{
f[x][j] = max(f[x][j], f[x][j - k] + f[y][k]);
}
}
}
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
int fa;
cin >> v[i] >> w[i] >> fa;
if (fa == -1)
root = i;
else
g[fa].push_back(i);
}
dfs(root);
cout << f[root][m];
return 0;
}
9.背包问题求方案数
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
const int mod = 1e9 + 7;
int f[N];
int cnt[N];
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
cnt[i] = 1; //什么都不装也是一种方案
}
for (int i = 1; i <= n; i++) {
int v, w;
scanf("%d%d", &v, &w);// 体积和价值
for (int j = m; j >= v; j--) {
int value = f[j - v] + w;
if (value > f[j]) {
f[j] = value;
cnt[j] = cnt[j - v];
}
else if (value == f[j]) {
cnt[j] = (cnt[j] + cnt[j - v]) % mod;
}
}
}
printf("%d", cnt[m]);
return 0;
}
10.背包问题求具体方案数
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1010;
int f[N][N];
int v[N], w[N];
int n, m;
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
cin >> v[i] >> w[i];
for (int i = n; i >= 1; i--)
{
for (int j = 0; j <= m; j++)
{
f[i][j] = f[i + 1][j];
if (j >= v[i])
f[i][j] = max(f[i][j], f[i + 1][j - v[i]] + w[i]);
}
}
int cur_v = m;
for (int i = 1; i <= n; i++)
{ //如果是最后一个元素，特判一下，防止越界即可
if (i == n && cur_v >= v[i])
{
printf("%d ", i);
break;
}
if (cur_v <= 0)
break;
//判断下标是否越界
if (cur_v - v[i] >= 0 && f[i][cur_v] == f[i + 1][cur_v - v[i]] + w[i]) {
printf("%d ", i);
cur_v = cur_v - v[i];//选了第i个物品，剩余容量就要减小。
}
}
return 0;
}
标签：混合背包 style back float tor 循环 max 优化 来源
原文地址：https://www.cnblogs.com/E-Dreamer-Blogs/p/12834300.html
