FZU 2214 Knapsack problem 【01背包】

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题目链接：https://cn.vjudge.net/problem/FZU-2214

思路：

背包的容量太大，dp 改成前 i个物品，价值为 v的容量是多少；（原本应该是前i个物品容量为 w的最大价值）现在改成v的最小容量；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
#include<vector>
#include<string>
#include<set>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define REP(i,n) for(int i=0;i<n;++i)

int read(){

    int r=0,f=1;char p=getchar();
    while(p>'9'||p<'0'){if(p=='-')f=-1;p=getchar();}
    while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}

typedef long long ll;
typedef unsigned long long ull;
const int Maxn = 5010;
const long long LINF = 1e18;
const int INF = 0x3f3f3f3f;
const int Mod = 10001;

ll dp[Maxn],w[Maxn],v[Maxn];

int main (void)
{
    int t,n,B;
    scanf("%d",&t);
    while (t--) {
        scanf("%d%d",&n,&B);
        for (int i = 0; i < n; ++i) scanf("%lld%lld",&w[i],&v[i]);
        for (int i = 0; i <= Maxn; ++i) dp[i] = INF;
        dp[0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = Maxn; j >= v[i]; --j) {  // 注意是从 Maxn到 v[i]，不是v[i]到Maxn更新数据
                if(dp[j-v[i]] >= INF) continue;
                dp[j] = min(dp[j],dp[j-v[i]]+w[i]);
            }
        }
        for (int i = Maxn; i >= 0; --i) {
                if(dp[i] <= B) {
                    printf("%d\n",i);
                    break;
                }
        }
    }
    return 0;
}