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来源：互联网发布：js的二维数组编辑：程序博客网时间：2024/05/18 03:46

dp[i][j][k]为在点(i,j)，当前获得量为k的方法是否存在

要求解点(i,j)，有(i-1,j)和(i,j-1)的答案即可得出(因为题目中只允许向下向右走

数据并不强，也并不需要压缩空间

bc上的题解是dp[i][j][k]代表在(i,j)背包容量为k时的最大值，个人认为反而复杂化了问题……

#include<iostream>#include<iomanip>#include<cstdio>#include<cstring>#include<sstream>#include<stack>#include<queue>#include<fstream>#include<algorithm>#include<map>#include<set>#include<vector>#include<cmath>#define CLR(x) memset(x,0,sizeof(x))#define SETMAX(x) memset(x,0x3f,sizeof(x))#define SETNO(x) memset(x,-1,sizeof(x))#define ll long long#define eps 1e-6#define pow2(x) ((x)*(x))#define forto(i,n) for(int i=0;i<n;i++)#define for1to(i,n) for(int i=1;i<=n;i++)#define VI vector<int>using namespace std;const double PI=acos(-1.0);#define INF 0x3f3f3f3f#define NINF 0xbfffffffusing namespace std;int A[111][111];bool dp[111][111][111];int main(){    cin.sync_with_stdio(false);    #ifndef ONLINE_JUDGE    freopen("test.txt","r",stdin);    #endif // ONLINE_JUDGE    int n,m,K;    while (cin>>n>>m>>K)    {        for1to(i,n)            for1to(j,m)                cin>>A[i][j];        for (int i=0;i<=n;i++)            dp[i][0][0]=true;        for (int i=0;i<=m;i++)            dp[0][i][0]=true;        for1to(i,n)        {            for1to(j,m)            {                for (int k=0;k<=K;k++)                {                    dp[i][j][k]=dp[i-1][j][k] || dp[i][j-1][k];                    if (k>=A[i][j])                        dp[i][j][k] = dp[i][j][k]|| dp[i-1][j][k-A[i][j]] || dp[i][j-1][k-A[i][j]];                }            }        }        for (int i=K;i>=0;i--)        {            if (dp[n][m][i])            {                cout<<i<<endl;                break;            }        }    }    return 0;}

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