Book 动态规划

 

虽然之前学过一点点，但是还是不会------现在好好跟着白书1.4节学一下——————

（1）数字三角形

d(i,j) = max(d(i+1,j),d(i+1,j+1)) + a[i][j]

hdu 2084

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 typedef long long LL;

14 const int INF = (1<<30)-1;

15 const int mod=1000000007;

16 const int maxn=1000005;

17 

18 int d[1005][1005],a[1005][1005];

19 

20 int main(){

21     int T;

22     scanf("%d",&T);

23     while(T--){

24         int n;

25         scanf("%d",&n);

26         memset(d,0,sizeof(d));

27         for(int i = 1;i<=n;i++)

28             for(int j = 1;j<=i;j++) scanf("%d",&a[i][j]);

29         

30         for(int i = n;i>=1;i--){

31             for(int j = 1;j<=i;j++)

32             d[i][j] = max(d[i+1][j],d[i+1][j+1]) + a[i][j];

33         }

34         printf("%d\n",d[1][1]);

35     }

36     return 0;

37 }

View Code

 

（2）嵌套矩形

把图先建出来，然后d(i) = max(d(j) + 1)

nyoj 16

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 typedef long long LL;

14 const int INF = (1<<30)-1;

15 const int mod=1000000007;

16 const int maxn=1000005;

17 

18 int g[1005][1005];

19 int d[1005];

20 int n;

21 

22 struct node{

23     int x,y;

24 }a[maxn];

25 

26 int dp(int i){

27     int& ans = d[i];

28     if(ans > 0) return ans;

29     ans = 1;

30     for(int j = 1;j <= n;j++) 

31     if(g[i][j]) ans = max(ans,dp(j) + 1);

32     

33     return ans;

34 }

35 

36 int work(){

37     int res = -1;

38     for(int i = 1;i <= n;i++) res = max(res,dp(i));

39     return res;

40 }

41 

42 int main(){

43     int T;

44     scanf("%d",&T);

45     while(T--){

46         scanf("%d",&n);

47         for(int i = 1;i <= n;i++) scanf("%d %d",&a[i].x,&a[i].y);

48         memset(g,0,sizeof(g));

49         memset(d,0,sizeof(d));

50         

51         for(int i = 1;i <= n;i++){

52             for(int j = 1;j <= n;j++){

53                 if((a[i].x > a[j].x && a[i].y > a[j].y) || (a[i].x > a[j].y && a[i].y > a[j].x ))

54                 g[i][j] = 1;

55             }

56         }

57         

58         printf("%d\n",work());

59     }

60     return 0;

61 }

View Code