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## 背包问题-最小价值-空间恰好$j$
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### 一、$01$背包
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求价值最小值：初始化$f[0][0] = 0$, 其余是$INF$
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例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品只能选一个，求 <font color='blue' size=4><b>总体积恰好是$m$</b></font>的 <font color='red' size=4><b>最小价值</b></font>
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输入
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```c++
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4 5
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1 2
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2 4
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3 4
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4 5
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```
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输出
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```c++
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7
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```
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#### 1、二维
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int INF = 0x3f3f3f3f;
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int n, m;
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int f[N][N];
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int main() {
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//文件输入
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freopen("QiaHao_01_Min.in", "r", stdin);
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scanf("%d %d", &n, &m);
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memset(f, 0x3f, sizeof f); //预求最小，先设最大
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f[0][0] = 0; //前0个物品中选择，装满0个空间，获得的最小值为0
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for (int i = 1; i <= n; i++) {
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int v, w;
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scanf("%d %d", &v, &w);
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for (int j = 0; j <= m; j++) {
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f[i][j] = f[i - 1][j];
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if (j >= v) f[i][j] = min(f[i][j], f[i - 1][j - v] + w);
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}
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}
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printf("%d\n", f[n][m]);
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return 0;
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}
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```
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#### 2、一维
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int INF = 0x3f3f3f3f;
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int n, m;
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int f[N];
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int main() {
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//文件输入
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freopen("QiaHao_01_Min.in", "r", stdin);
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scanf("%d %d", &n, &m);
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memset(f, 0x3f, sizeof f);
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f[0] = 0;
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for (int i = 1; i <= n; i++) {
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int v, w;
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scanf("%d %d", &v, &w);
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for (int j = m; j >= v; j--)
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f[j] = min(f[j], f[j - v] + w);
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}
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printf("%d\n", f[m]);
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return 0;
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}
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```
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### 二、完全背包
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求价值最小值：初始化$f[0][0] = 0$, 其余是$INF$
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例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品可以选无数多个，求 <font color='blue' size=4><b>总体积恰好是$m$</b></font>的 <font color='red' size=4><b>最小价值</b></font>
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输入
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```c++
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4 5
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1 2
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2 4
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3 4
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4 5
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```
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输出
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```c++
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7
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```
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#### 1、二维
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int INF = 0x3f3f3f3f;
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int n, m;
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int f[N][N];
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int main() {
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scanf("%d %d", &n, &m);
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memset(f, 0x3f, sizeof f);
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f[0][0] = 0;
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for (int i = 1; i <= n; i++) {
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int v, w;
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scanf("%d %d", &v, &w);
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for (int j = 0; j <= m; j++) {
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f[i][j] = f[i - 1][j];
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if (j >= v) f[i][j] = min(f[i][j], f[i][j - v] + w);
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}
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}
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printf("%d\n", f[n][m]);
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return 0;
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}
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```
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#### 2、一维
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110, INF = 0x3f3f3f3f;
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int n, m;
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int f[N];
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int main() {
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//文件输入
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freopen("QiaHao_WQ_Min.in", "r", stdin);
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scanf("%d %d", &n, &m);
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memset(f, 0x3f, sizeof f);
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f[0] = 0;
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for (int i = 1; i <= n; i++) {
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int v, w;
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scanf("%d %d", &v, &w);
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for (int j = v; j <= m; j++)
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f[j] = min(f[j], f[j - v] + w);
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}
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printf("%d\n", f[m]);
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return 0;
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}
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```
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