这道题的DP的状态设计的很有想法啊。
假如我们一行一行来选择的话，状态将会极其复杂。
如果一列一列来看的话，比如你想选aij,那么第i列的前j个都要选，并且第i+1列的前j-1个都要选。
于是状态就很好设计了，定义dp[n][i][j]表示还剩下n个要选的砖块，当前选择第i列的前j个所能达到的最大分值。
那么dp[n][i][j]=max(dp[n-j][i+1][k]+sum[i][j])(j-1<=k<=n-i).
记忆化搜索一下就OK了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-7 # define MOD 1024523 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=55; //Code begin... int a[N][N], sum[N][N], dp[N*N][N][N], n; int dfs(int x, int col, int row){ if (x<0||(col>n&&x)) return -INF; if (~dp[x][col][row]) return dp[x][col][row]; if (x==0) return row==0?0:-INF; int res=-INF; FOR(i,max(0,row-1),min(x,n-col)) res=max(res,dfs(x-row,col+1,i)+sum[col][row]); return dp[x][col][row]=res; } int main () { int m, ans=0; scanf("%d%d",&n,&m); mem(dp,-1); FOR(i,1,n) FOR(j,1,n-i+1) scanf("%d",&a[i][j]), sum[j][i]=sum[j][i-1]+a[i][j]; FOR(i,0,n) ans=max(ans,dfs(m,1,i)); printf("%d ",ans); return 0; }