第一题:


AC代码:
(1)空间复杂度为O(n^2)的算法:

#include<bits/stdc++.h>
using namespace std;
const int N = 150;
int dp[N][N],a[N][N];

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		memset(dp,0,sizeof(dp));
		int r,c;
		scanf("%d%d",&r,&c);
		for(int i=1;i<=r;i++){
			for(int j=1;j<=c;j++){
				scanf("%d",&a[i][j]);
				dp[i][j]=max(dp[i-1][j],dp[i][j-1])+a[i][j];
			}
		}
		printf("%d\n",dp[r][c]);
	}
} 

(2)空间复杂度为O(2n)的算法:

#include<bits/stdc++.h>
using namespace std;
const int N = 150;
int dp[2][N],a[N][N];

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		memset(dp,0,sizeof(dp));
		int r,c;
		scanf("%d%d",&r,&c);
		for(int i=1;i<=r;i++){
			for(int j=1;j<=c;j++){
				scanf("%d",&a[i][j]);
				dp[i&1][j]=max(dp[(i&1)^1][j],dp[i&1][j-1])+a[i][j];
			}
		}
		printf("%d\n",dp[r&1][c]);
	}
} 

(3)空间复杂度为O(n)的算法:

#include<bits/stdc++.h>
using namespace std;
const int N = 150;
int dp[N],a[N][N];

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		memset(dp,0,sizeof(dp));
		int r,c;
		scanf("%d%d",&r,&c);
		for(int i=1;i<=r;i++){
			for(int j=1;j<=c;j++){
				scanf("%d",&a[i][j]);
				dp[j]=max(dp[j],dp[j-1])+a[i][j];
			}
		}
		printf("%d\n",dp[c]);
	}
} 

第二题:


AC代码:

#include<bits/stdc++.h>
using namespace std;
const int N = 150;
int dp[N],a[N][N];

int main(){
	
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			scanf("%d",&a[i][j]);
			if(i==1)dp[j]=dp[j-1]+a[i][j];
			else if(j==1)dp[j]+=a[i][j];
			else dp[j]=min(dp[j],dp[j-1])+a[i][j];
		}
	}
	printf("%d\n",dp[n]);
} 

第三题:


解析:
由于当路径重合时i1+j1=i2+j2=k,所以我们可以用k和i表示出一个j
那么分为四种情况:
p1:(i1,j1) p2:(i2,j2)
p1右p2右:(k,i1,i2)
p1下p2右:(k,i1+1,i2)
p1右p2下:(k,i1,i2+1)
p1下p2下:(k,i1+1,i2+1)
由于1<=x<=n&&1<=k-x<=n可以推出x<=k-1&&x>=k-n
AC代码:

#include<bits/stdc++.h>
using namespace std;
const int N = 20;
int dp[N][N][N],a[N][N];

int main(){
	
	int n;
	scanf("%d",&n);
	int r,c,num;
	while(scanf("%d%d%d",&r,&c,&num)){
		if(r==0&&c==0&&num==0)break;
		a[r][c]=num;
	}
	for(int k=1;k<=n+n;k++){
		for(int i1=max(1,k-n);i1<=k-1;i1++){
			for(int i2=max(1,k-n);i2<=k-1;i2++){
				int j1=k-i1,j2=k-i2;
				if(j1>=1&&j1<=n&&j2>=1&&j2<=n){
					int t = a[i1][j1];
					if(i1!=i2)t+=a[i2][j2];
					int &x=dp[k][i1][i2];
					x=max(x,dp[k-1][i1][i2]+t);
					x=max(x,dp[k-1][i1-1][i2]+t);
					x=max(x,dp[k-1][i1][i2-1]+t);
					x=max(x,dp[k-1][i1-1][i2-1]+t);
				}
			}
		}
	}
	printf("%d\n",dp[n+n][n][n]);
} 
 

第四题:
AC代码:

#include<bits/stdc++.h>
using namespace std;
const int N = 55;
int dp[2*N][N][N],a[N][N];

int main(){
	
	int n,m;
	scanf("%d%d",&n,&m);
	int r,c,num;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			scanf("%d",&a[i][j]);
		}
	}
	for(int k=1;k<=n+m;k++){
		for(int i1=max(1,k-m);i1<=min(k-1,n);i1++){
			for(int i2=max(1,k-m);i2<=min(k-1,n);i2++){
				int j1=k-i1,j2=k-i2;
				if(j1>=1&&j1<=m&&j2>=1&&j2<=m){
					int t = a[i1][j1];
					if(i1!=i2)t+=a[i2][j2];
					int &x=dp[k][i1][i2];
					x=max(x,dp[k-1][i1][i2]+t);
					x=max(x,dp[k-1][i1-1][i2]+t);
					x=max(x,dp[k-1][i1][i2-1]+t);
					x=max(x,dp[k-1][i1-1][i2-1]+t);
				}
			}
		}
	}
	printf("%d\n",dp[n+m][n][n]);
}