CSP2020-j2 T4 方格取数

题目链接：https://www.luogu.com.cn/problem/P7074

一、dfs(25分)没有任何优化，时间复杂度约（O（3^(n*m)））

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n, m;
 4 int a[1005][1005], vis[1005][1005];
 5 int nex[3][2]={{-1,0},{1,0},{0,1}};
 6 int sum;
 7 int ans=-0x7fffffff;
 8 void dfs(int x, int y){
 9     if(x==1 && y==1){
10         vis[x][y]=1;
11         sum=a[x][y];
12     }
13     if(x==n && y==m){
14         ans=max(ans, sum);
15         return;
16     }
17     for(int i=0; i<3; i++){
18         int nx=x+nex[i][0], ny=y+nex[i][1];
19         if(nx<1 || nx>n || ny<1 || ny>m)continue;
20         if(vis[nx][ny])continue;
21         vis[nx][ny]=1;
22         sum+=a[nx][ny];
23         dfs(nx, ny);
24         sum-=a[nx][ny];
25         vis[nx][ny]=0;
26     }
27 }
28 int main()
29 {
30     cin>>n>>m;
31     for(int i=1; i<=n; i++)
32         for(int j=1; j<=m; j++)
33             cin>>a[i][j];                
34     dfs(1,1);
35     cout<<ans;
36     return 0;
37  }

 

同名题：P1004 [NOIP2000 提高组] 方格取数

https://www.luogu.com.cn/problem/P1004

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n, a[15][15];
 4 int f[15][15][15][15];
 5 int dfs(int x1, int y1, int x2, int y2){
 6     if(f[x1][y1][x2][y2]!=-1)return f[x1][y1][x2][y2];
 7     if(x1==n && y1==n && x2==n && y2==n)return 0;
 8     int ret=0;
 9     if(x1 < n && x2 < n) 
10         ret = max(ret, dfs(x1+1, y1, x2+1, y2) + a[x1+1][y1] + a[x2+1][y2] - a[x1+1][y1] * (x1+1 == x2+1 && y1 == y2)); 
11     if(x1 < n && y2 < n)
12         ret = max(ret, dfs(x1+1, y1, x2, y2+1) + a[x1+1][y1] + a[x2][y2+1] - a[x1+1][y1] * (x1+1 == x2 && y1 == y2+1));
13     if(y1 < n && x2 < n)
14         ret = max(ret, dfs(x1, y1+1, x2+1, y2) + a[x1][y1+1] + a[x2+1][y2] - a[x1][y1+1] * (x1 == x2+1 && y1+1 == y2));
15     if(y1 < n && y2 < n) 
16         ret = max(ret, dfs(x1, y1+1, x2, y2+1) + a[x1][y1+1] + a[x2][y2+1] - a[x1][y1+1] * (x1 == x2 && y1+1 == y2+1)); 
17     return f[x1][y1][x2][y2]=ret;
18     
19 }
20 int main()
21 {
22     scanf("%d", &n);
23     int x, y, z;
24     while(1){
25         scanf("%d%d%d", &x, &y, &z);
26         if(x==0 && y==0 && z==0)break;
27         a[x][y]=z;
28     }
29     memset(f, -1, sizeof(f));
30     printf("%d", dfs(1, 1, 1, 1)+ a[1][1]);
31     return 0;
32 }

 

posted @ 2020-11-19 15:20  TFLSNOI  阅读(579)  评论(0)    收藏  举报