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  • DP录 （更新）

    补补弱项 全面发展。。

    从最基础来

    sdut1299最长上升子序

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[1100];
 8 int a[1100];
 9 int main()
10 {
11     int i,j,k,n;
12     cin>>n;
13     for(i =1; i <= n ; i++)
14     {
15         cin>>a[i];
16         dp[i] = 1;
17     }
18     for(i = 1; i <= n ;i++)
19     for(j = 1 ; j < i ; j++)
20     if(a[i]>a[j])
21     {
22         dp[i] = max(dp[j]+1,dp[i]);
23     }
24     int maxz = 0;
25     for(i = 1; i <= n ;i++)
26     maxz = max(dp[i],maxz);
27     cout<<maxz<<endl;
28     return 0;
29 }
30  
31 
32 
33 
34 /**************************************
35     Problem id    : SDUT OJ 1299 
36     User name    : 尚雨 
37     Result        : Accepted 
38     Take Memory    : 476K 
39     Take Time    : 0MS 
40     Submit Time    : 2013-08-05 09:26:48  
41 **************************************/
    View Code

    hdu2084数塔

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[110][110];
 8 int a[110][110];
 9 int main()
10 {
11     int i,j,k,n,t;
12     cin>>t;
13     while(t--)
14     {
15         cin>>n;
16         memset(dp,0,sizeof(dp));
17         for(i = 1; i <= n ;i++)
18         for(j = 1; j <= i ;j++)
19         cin>>a[i][j];
20         for(j = 1; j <= n ; j++)
21         dp[n][j] = a[n][j];
22         for(i = n-1 ; i >= 1 ;i--)
23         for(j = 1; j <= i ; j++)
24         {
25             dp[i][j]=a[i][j]+max(dp[i+1][j],dp[i+1][j+1]);
26         }
27         cout<<dp[1][1]<<endl;
28     }
29     return 0;
30 }
    View Code

    hdu1159最长公共子序列

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[1100][1100];
 8 char s1[1100],s2[1100];
 9 int main()
10 {
11     int i,j,k1,k2;
12     while(cin>>s1)
13     {
14         k1 = strlen(s1);
15         cin>>s2;
16         k2 = strlen(s2);
17         memset(dp,0,sizeof(dp));
18         for(i = 1 ; i <= k1 ; i++)
19             for(j = 1 ; j <= k2 ; j++)
20             {
21                 if(s1[i-1]==s2[j-1])
22                 dp[i][j] = dp[i-1][j-1]+1;
23                 else
24                 dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
25             }
26         cout<<dp[k1][k2]<<endl;
27     }
28     return 0;
29 }
    View Code

    hdu1003最大子段和

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define INF 0xfffffff
 8 int dp[110000];
 9 int a[100010];
10 int main()
11 {
12     int i,j,n,t,kk=0;
13     cin>>t;
14     while(t--)
15     {
16         cin>>n;kk++;
17         for(i = 1; i <= n ;i++)
18         cin>>a[i];
19         int maxz = -INF,tmaxz=0;
20         int x=1,y,tx=1;
21         for(i =  1; i <= n ;i++)
22         {
23             tmaxz+=a[i];
24             if(tmaxz>maxz)
25             {
26                 maxz = tmaxz;
27                 x = tx;
28                 y = i;
29             }
30             if(tmaxz<0)
31             {
32                 tmaxz=0;
33                 tx = i+1;
34             }
35         }
36         printf("Case %d:
",kk);
37         cout<<maxz<<" "<<x<<" "<<y<<endl;
38         if(t!=0)
39         puts("");
40     }
41     return 0;
42 }
    View Code

     sdut1225http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1225

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 char s1[1050],s2[1050];
 8 int dp[1050][1050];
 9 int main()
10 {
11     int i,j,k1,k2;
12     while(cin>>s1)
13     {
14         cin>>s2;
15         memset(dp,0,sizeof(dp));
16         k1 = strlen(s1);
17         k2 = strlen(s2);
18         for(i = 0 ; i <= k1 ; i++)
19         dp[i][0] = i;
20         for(i = 0 ; i <= k2 ; i++)
21         dp[0][i] = i;
22         for(i = 1 ; i <= k1 ; i++)
23             for(j = 1 ; j <= k2 ; j++)
24             {
25                 if(s1[i-1]==s2[j-1])
26                 dp[i][j] = dp[i-1][j-1];
27                 else
28                 dp[i][j] = dp[i-1][j-1]+1;
29                 dp[i][j] = min(dp[i][j],min(dp[i-1][j]+1,dp[i][j-1]+1));
30             }
31         cout<<dp[k1][k2]<<endl;
32     }
33     return 0;
34 }
    View Code

     hdu1058 预处理出所有的数 预处理的技巧挺好

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 #include<algorithm>
 6 using namespace std;
 7 int dp[6010];
 8 void init()
 9 {
10     int i,j,a=2,b=3,c=5,d=7,ii,jj,g,o;
11     ii=jj=g=o=1;
12     dp[1] = 1;
13     for(i = 2; i <= 5842 ;i++)
14     {
15         int x = dp[ii]*a;
16         int y = dp[jj]*b;
17         int z = dp[g]*c;
18         int w = dp[o]*d;
19         dp[i] = min(min(x,y),min(z,w));
20         if(dp[i]==x)
21         ii++;
22         if(dp[i]==y)
23         jj++;
24         if(dp[i]==z)
25         g++;
26         if(dp[i]==w)
27         o++;
28     }
29 }
30 int main()
31 {
32     int i,j,k,n;
33     init();
34     while(cin>>n)
35     {
36         if(n==0)
37         break;
38         if(n%10==1&&n%100!=11)
39         printf("The %dst humble number is %d.
",n,dp[n]);
40         else if(n%10==2&&n%100!=12)
41         printf("The %dnd humble number is %d.
",n,dp[n]);
42         else if(n%10==3&&n%100!=13)
43         printf("The %drd humble number is %d.
",n,dp[n]);
44         else
45         printf("The %dth humble number is %d.
",n,dp[n]);
46     }
47     return 0;
48 }
    View Code

    hdu2571 类似数塔

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define INF 0xfffffff
 6 using namespace std;
 7 int dp[30][1010];
 8 int v[30][1010];
 9 int main()
10 {
11     int i,j,n,m,t,g;
12     cin>>t;
13     while(t--)
14     {
15         cin>>n>>m;
16         for(i = 1; i <= n ;i++)
17             for(j = 1; j <= m ; j++)
18             {
19                cin>>v[i][j];
20                dp[i][j] = -INF;
21             }
22         dp[1][1] = v[1][1];
23         for(i = 1;i <= n ;i++)
24             for(j = 1; j <= m ; j++)
25             {
26                 if(i<n)
27                 dp[i+1][j] = max(dp[i][j]+v[i+1][j],dp[i+1][j]);
28                 if(j<m)
29                 dp[i][j+1] = max(dp[i][j]+v[i][j+1],dp[i][j+1]);
30                 for(g = 2;;g++)
31                 {
32                     if(g*j>m)
33                     break;
34                     dp[i][g*j] = max(dp[i][j]+v[i][g*j],dp[i][g*j]);
35                 }
36             }
37         cout<<dp[n][m]<<endl;
38     }
39     return 0;
40 }
    View Code

     嵌套矩形 类似最长上升子序 因没排序WA了几次

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 struct node
 8 {
 9     int a,b;
10 }q[1010];
11 int dp[1010];
12 bool cmp(node a,node b)
13 {
14     if(a.a==b.a)
15     return a.b<b.b;
16     return a.a<b.a;
17 }
18 int main()
19 {
20     int i,j,t,n,tt;
21     cin>>t;
22     while(t--)
23     {
24         cin>>n;
25         memset(dp,0,sizeof(dp));
26         for(i = 1; i <= n ;i++)
27         {
28             cin>>q[i].a>>q[i].b;
29             if(q[i].a>q[i].b)
30             {
31                 tt = q[i].a;
32                 q[i].a = q[i].b;
33                 q[i].b = tt;
34             }
35             dp[i] = 1;
36         }
37         sort(q+1,q+n+1,cmp);
38         for(i = 1; i <= n ; i++)
39             for(j = 1; j  < i ; j++)
40             {
41                 if(((q[i].a>q[j].a)&&(q[i].b>q[j].b))||(q[i].b>q[j].a&&q[i].a>q[j].b))
42                 dp[i] = max(dp[i],dp[j]+1);
43             }
44         int maxz=0;
45         for(i = 1; i <= n ;i++)
46         maxz = max(maxz,dp[i]);
47         cout<<maxz<<endl;
48     }
49     return 0;
50 }
    View Code

        Vijos 1493 传纸条 多进程DP

    四维

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[51][51][51][51];
 8 int a[51][51];
 9 int main()
10 {
11     int i,j,n,m,g,o;
12     while(cin>>n>>m)
13     {
14         memset(dp,0,sizeof(dp));
15         memset(a,0,sizeof(a));
16         for(i = 1; i <= n ; i++)
17             for(j = 1; j <= m ; j++)
18             scanf("%d",&a[i][j]);
19         for(i = 1; i <= n ;i++)
20             for(j = 1 ; j <= m ; j++)
21                 for(g = 1; g <= n ; g++)
22                     for(o = 1 ;o <= m ;o++)
23                     {
24                         dp[i][j][g][o] = max(max(dp[i-1][j][g-1][o],dp[i-1][j][g][o-1]),max(dp[i][j-1][g-1][o],dp[i][j-1][g][o-1]));
25                         if(i==g&&j==o)
26                         dp[i][j][g][o]+=a[i][j];
27                         else
28                         dp[i][j][g][o]+=a[i][j]+a[g][o];
29                     }
30         printf("%d
",dp[n][m][n][m]);
31     }
32     return 0;
33 }
    View Code

    因为它只能从两个方向走过来 很巧妙的可以转换为3维的 

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[101][51][51];
 8 int a[51][51],d[101][51];
 9 int main()
10 {
11     int i,j,n,m,g,o,x,y,z,p;
12     while(cin>>n>>m)
13     {
14         memset(dp,0,sizeof(dp));
15         memset(d,0,sizeof(d));
16         for(i = 1; i <= n ; i++)
17             for(j = 1; j <= m ; j++)
18             {
19                 scanf("%d",&a[i][j]);
20                 if(i+j>n)
21                 d[i+j-1][n-i+1] = a[i][j];
22                 else
23                 d[i+j-1][j] = a[i][j];
24             }
25         for(i = 1; i < n+m ; i++)
26             for(j = 1 ;j <= m ; j++)
27                 for(g = 1; g <= m ;g++)
28                 {
29                     if(i <= n)
30                     {
31                         x = dp[i-1][j][g-1];
32                         y = dp[i-1][j-1][g-1];
33                         z = dp[i-1][j][g];
34                         p = dp[i-1][j-1][g];
35                     }
36                     else
37                     {
38                         x = dp[i-1][j][g];
39                         y = dp[i-1][j+1][g+1];
40                         z = dp[i-1][j+1][g];
41                         p = dp[i-1][j][g+1];
42                     }
43                     dp[i][j][g] = max(max(x,y),max(z,p));
44                     if(j==g)
45                     dp[i][j][g]+=d[i][j];
46                     else
47                     dp[i][j][g]+=d[i][j]+d[i][g];
48                 }
49         cout<<dp[n+m-1][1][1]<<endl;
50 
51     }
52     return 0;
53 }
    View Code

     hdu1059http://acm.hdu.edu.cn/showproblem.php?pid=1059

    多重背包 优化一下 时间卡的很紧

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<stdlib.h>
 7 using namespace std;
 8 int a[10],b[20010],d[33];
 9 int dp[1000010];
10 int main()
11 {
12     int i,j,k,s=0,g,n=0,kk=0;
13     while(scanf("%d",&a[1])!=EOF)
14     {
15         s=0;
16         s+=a[1];g=0;kk++;
17         memset(b,0,sizeof(b));
18         for(i = 2; i <= 6 ; i++)
19         {
20             scanf("%d",&a[i]);
21             s+=a[i]*i;
22             n+=a[i];
23         }
24         if(s==0) break;
25         printf("Collection #%d:
",kk);
26         if(s%2!=0)
27         {
28             printf("Can't be divided.

");
29             continue;
30         }
31         else
32         s = s/2;
33         d[1] = 1;
34         for(i = 0 ; i <= s ; i++)
35         dp[i] = 0;
36         for(i = 2; i  < 30 ; i++)
37         d[i] = d[i-1]<<1;
38         j=0;
39         for(i = 1; i <= 6 ; i++)
40         {
41             k = 1;
42             while(a[i]>d[k])
43             {
44                 a[i]-=d[k];
45                 b[++j] = d[k]*i;
46                 k++;
47             }
48             if(a[i])
49             {
50                 b[++j] = a[i]*i;
51             }
52         }
53         k = j;
54         for(i = 1; i <= k ; i++)
55         for(j = s ; j >= b[i] ; j--)
56         {
57             dp[j] = max(dp[j-b[i]]+b[i],dp[j]);
58         }
59         if(dp[s]==s)
60         printf("Can be divided.
");
61         else
62         printf("Can't be divided.
");
63         puts("");
64     }
65     return 0;
66 }
    View Code

     hdu1176免费馅饼 http://acm.hdu.edu.cn/showproblem.php?pid=1176

    将每秒的状态合起来类似数塔 倒推回去就行

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[100010][15];
 8 int main()
 9 {
10     int i,j,k,n,x,t;
11     while(scanf("%d",&n)&&n)
12     {
13         memset(dp,0,sizeof(dp));
14         int maxz=0;
15         for(i = 1 ;i <= n ; i++)
16         {
17             scanf("%d%d",&x,&t);
18             dp[t][x+1]++;
19             maxz = max(maxz,t);
20         }
21         for(i = maxz-1 ; i >= 0; i--)
22            for(j = 1 ; j <= 11 ; j++)
23            {
24                dp[i][j]+=max(max(dp[i+1][j-1],dp[i+1][j+1]),dp[i+1][j]);
25            }
26         cout<<dp[0][6]<<endl;
27     }
28     return 0;
29 }
    View Code

     hdu1069http://acm.hdu.edu.cn/showproblem.php?pid=1069 每个矩形算六个 排序后类似最长上升子序

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #define N 40
 7 using namespace std;
 8 struct node
 9 {
10     int x,y,z;
11 }rc[N*6];
12 int g,dp[310];
13 void init(int a,int b,int c)
14 {
15     rc[++g].x = a;
16     rc[g].y = b;
17     rc[g].z = c;
18     rc[++g].x = b;
19     rc[g].y = a;
20     rc[g].z = c;
21     rc[++g].x = a;
22     rc[g].y = c;
23     rc[g].z = b;
24     rc[++g].x = b;
25     rc[g].y = c;
26     rc[g].z = a;
27     rc[++g].x = c;
28     rc[g].y = b;
29     rc[g].z = a;
30     rc[++g].x = c;
31     rc[g].y = a;
32     rc[g].z = b;
33 }
34 bool cmp(node a,node b)
35 {
36     if(a.x==b.x)
37     return a.y>b.y;
38     return a.x>b.x;
39 }
40 int main()
41 {
42     int i,j,n,a,b,c,kk=0;
43     while(cin>>n)
44     {
45         g=0;kk++;
46         if(n==0) break;
47         memset(dp,0,sizeof(dp));
48         for(i = 1;i <= n ; i++)
49         {
50             scanf("%d%d%d",&a,&b,&c);
51             init(a,b,c);
52         }
53         sort(rc+1,rc+g+1,cmp);
54         for(i = 1; i <= g ; i++)
55         dp[i] = rc[i].z;
56         for(i = 1; i <= g ; i++)
57             for(j = 1; j < i ; j++)
58             {
59                 if(rc[j].x>rc[i].x&&rc[j].y>rc[i].y)
60                 dp[i] = max(dp[i],dp[j]+rc[i].z);
61             }
62         int maxz=0;
63         printf("Case %d: maximum height = ",kk);
64         for(i = 1; i <= g ; i++)
65         maxz = max(dp[i],maxz);
66         cout<<maxz<<endl;
67     }
68     return 0;
69 }
    View Code

     hdu1081http://acm.hdu.edu.cn/showproblem.php?pid=1081 记录从i到j行g列的和 转换为一维的最大子段和 

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 #include<algorithm>
 6 using namespace std;
 7 #define INF 0xfffffff
 8 int a[110][110],s[105][105][105],c[105][105][105];
 9 int main()
10 {
11     int i,j,k,n,g;
12     while(scanf("%d",&n)!=EOF)
13     {
14         memset(s,0,sizeof(s));
15         memset(c,0,sizeof(c));
16         for(i = 1; i <= n ; i++)
17             for(j = 1 ;j <= n ;j++)
18             scanf("%d",&a[i][j]);
19         for(i = 1; i <= n ; i++)
20             for(j = 1 ;j <= n ; j++)
21             {
22                 s[j][j][i] = a[j][i];
23                 for(g = j+1; g <= n ; g++)
24                 s[j][g][i]=a[g][i]+s[j][g-1][i];
25             }
26         int maxz = -INF,tmaxz=0;
27         for(i = 1; i <= n ; i++)
28             for(j = 1; j <= n ; j++)
29             {
30                 tmaxz=0;
31                 for(g = 1 ;g <= n ;g++)
32                 {
33                     tmaxz+=s[i][j][g];
34                     if(tmaxz>maxz)
35                     maxz=tmaxz;
36                     if(tmaxz<0)
37                     tmaxz = 0;
38                 }
39             }
40           cout<<maxz<<endl;
41 
42     }
43     return 0;
44 }
    View Code

     hdu1087 水题

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define LL long long
 8 int a[1010];
 9 LL dp[1010];
10 int main()
11 {
12     int i,j,k,n;
13     while(cin>>n)
14     {
15         if(!n) break;
16         memset(dp,0,sizeof(dp));
17         for(i = 1; i <= n ; i++)
18         {
19             cin>>a[i];
20             dp[i] = a[i];
21         }
22         for(i = 1 ;i <= n ; i++)
23             for(j = 1; j < i ; j++)
24             if(a[i]>a[j])
25             dp[i] = max(dp[i],dp[j]+a[i]);
26         LL maxz=dp[1];
27         for(i = 2 ;i <= n ;i++)
28         if(dp[i]>maxz)
29         maxz = dp[i];
30         cout<<maxz<<endl;
31     }
32     return 0;
33 }
    View Code

    hdu1158http://acm.hdu.edu.cn/showproblem.php?pid=1158二维DP 三重循环 以前居然做过 好扯 居然一点印象没有 好认真的推了半张纸。。

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define INF 0xfffffff
 8 int dp[20][1010],num[20];
 9 int main()
10 {
11     int i,j,k,n,m,a,b,c,g;
12     while(cin>>n)
13     {
14         if(!n) break;
15         cin>>a>>b>>c;
16         memset(dp,0,sizeof(dp));
17         memset(num,0,sizeof(num));
18         int maxz=0,minz;
19         for(i = 1; i <= n ;i++)
20         {
21             cin>>num[i];
22             if(num[i]>maxz)
23             maxz = num[i];
24             if(num[i]<minz)
25             minz = num[i];
26         }
27         for(i = 1;i <= n ;i++)
28             for(j = 0 ; j <= maxz ; j++)
29             dp[i][j] = INF;
30         for(i = 0 ; i <= maxz ; i++)
31         dp[1][i] = i*(a+b);
32         for(i = 2; i <= n ; i++)
33             for(j = num[i-1] ; j <= maxz ; j++)
34                 for(g = num[i] ; g <= maxz ; g++)
35                 {
36                     if(g>j)
37                     dp[i][g] = min(dp[i][g],dp[i-1][j]+(g-j)*a+g*b);
38                     else if(g<j)
39                     dp[i][g] = min(dp[i][g],dp[i-1][j]+(j-g)*c+g*b);
40                     else
41                     dp[i][g] = min(dp[i][g],dp[i-1][j]+g*b);
42                 }
43         int maxzz = INF;
44         for(i = num[n] ;i <= maxz ;i++)
45         if(dp[n][i]<maxzz)
46           maxzz = dp[n][i];
47         cout<<maxzz<<endl;
48     }
49     return 0;
50 }
    View Code

     hdu1165http://acm.hdu.edu.cn/showproblem.php?pid=1165 这题确实不能算DP  我还傻乎乎的写了一递归  M比较小 直接推就好 了

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define LL long long
 8 #define N 1000010
 9 int n,m;
10 LL dp[4][N+10];
11 int main()
12 {
13     int i,j;
14     for(i = 0 ; i <= N ; i++)
15     dp[0][i] = i+1;
16     dp[1][0] = 2;
17     dp[2][0] = 3;
18     dp[3][0] = 5;
19     for(i = 1 ; i <= N ; i++)
20     dp[3][i] = dp[1][0]*dp[3][i-1]+3;
21     while(scanf("%d%d",&m,&n)!=EOF)
22     {
23         if(m==1)
24         dp[m][n] = dp[1][0]+n;
25         else if(m==2)
26         dp[m][n] = dp[1][0]*n+dp[2][0];
27         cout<<dp[m][n]<<endl;
28     }
29     return 0;
30 }
    View Code

     hdu1506 http://acm.hdu.edu.cn/status.php 题不错 记录i左边及右边第一个比它小的位置 用来求以它为高的最大面积 求位置的时候 dp一下下吧 

    注意。。杭电上long long会无限WA。。用——int64

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define N 100100
 8 __int64 a[N],dp1[N],dp2[N];
 9 __int64 s;
10 int main()
11 {
12     int i,n;
13     while(cin>>n)
14     {
15         if(!n) break;
16         for(i = 1 ;i <= n ;i++)
17         scanf("%I64d",&a[i]);
18         dp1[1] = 0;
19         a[0] = -1;
20         for(i = 2; i <= n ;i++)
21         {
22             if(a[i]>a[i-1])
23             dp1[i] = i-1;
24             else
25             {
26                 int x = i-1;
27                 while(x>=1&&a[x]>=a[i])
28                 {
29                     x = dp1[x];
30                 }
31                 dp1[i] = x;
32             }
33         }
34         dp2[n] = n+1;
35         a[n+1] = -1;
36         for(i = n-1; i >= 1 ;i--)
37         if(a[i]>a[i+1])
38         dp2[i] = i+1;
39         else
40         {
41             int x = i+1;
42             while(x<=n&&a[x]>=a[i])
43             {
44                 x = dp2[x];
45             }
46             dp2[i] = x;
47         }
48         long long maxz=0;
49         for(i = 1 ;i <= n ; i++)
50         {
51             s = (dp2[i]-dp1[i]-1)*a[i];
52             maxz = max(s,maxz);
53         }
54         printf("%I64d
",maxz);
55     }
56     return 0;
57 }
    View Code

     hdu1712http://acm.hdu.edu.cn/showproblem.php?pid=1712 简单二维DP

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int dp[110][110];
 8 int a[110][110];
 9 int main()
10 {
11     int i,j,g,n,m;
12     while(cin>>n>>m)
13     {
14         if(!n&&!m) break;
15         memset(a,0,sizeof(a));
16         memset(dp,0,sizeof(dp));
17         for(i = 1;i <= n ; i++)
18             for(j = 1 ;j <= m ; j++)
19             scanf("%d",&a[i][j]);
20         for(i = 1;i <= n ; i++)
21             for(j = 1 ; j <= m ; j++)
22             for(g = 0; g<= j ; g++)
23             dp[i][j] = max(dp[i-1][g]+a[i][j-g],dp[i][j]);
24         int maxz = 0;
25         for(i = 0; i <= m ;i++)
26         maxz = max(maxz,dp[n][i]);
27         cout<<maxz<<endl;
28     }
29     return 0;
30 }
    View Code

     hdu2372http://acm.hdu.edu.cn/showproblem.php?pid=2372 又一水DP

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define LL __int64
 8 LL dp[110][110];
 9 int a[110];
10 int main()
11 {
12     int i,j,g,n,k;
13     while(cin>>n>>k)
14     {
15         if(!n&&!k) break;
16         memset(dp,0,sizeof(dp));
17         for(i =1 ;i <= n ; i++)
18         {
19             scanf("%d",&a[i]);
20             dp[i][1] = 1;
21         }
22         for(i = 1; i <= n ;i++)
23             for(j = 1;j<=i&&j <= k ; j++)
24             for(g = 1 ; g < i ; g++)
25             if(a[i]>a[g])
26             dp[i][j] += dp[g][j-1];
27         LL maxz=0;
28         for(i = k ; i <= n ; i++)
29             maxz+=dp[i][k];
30         cout<<maxz<<endl;
31     }
32     return 0;
33 }
    View Code

     hdu4223http://acm.hdu.edu.cn/showproblem.php?pid=4223 这题不能算DP吧 预处理一下

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<cmath>
 7 using namespace std;
 8 #define LL __int64
 9 #define INF 0xfffffff
10 int s[1010];
11 int a[1010];
12 int main()
13 {
14     int i,j,k,n,kk=0,t;
15     cin>>t;
16     while(t--)
17     {
18         cin>>n;kk++;
19         memset(s,0,sizeof(s));
20         for(i = 1; i <= n ; i++)
21         {
22             scanf("%d",&a[i]);
23             s[i] = s[i-1]+a[i];
24         }
25         int minz=INF;
26         for(i = 1; i <= n ; i++)
27             for(j = 0; j < i ; j++)
28             {
29                 int x = abs(s[i]-s[j]);
30                 minz = min(minz,x);
31             }
32         printf("Case %d: ",kk);
33         cout<<minz<<endl;
34     }
35     return 0;
36 }
    View Code

     hdu4159http://acm.hdu.edu.cn/showproblem.php?pid=4159

    一直想推概率 结果怎么也推不出来 后来想到可以推可能的种数  因为太大WA了 又去想推概率 还是没退出 最后除了个很大的数 过了

     1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 #include<algorithm>
 6 using namespace std;
 7 #define MM 1000000000000000.0
 8 double dp[55][55];
 9 int main()
10 {
11     int n,s,i,j;
12     while(cin>>n>>s)
13     {
14         memset(dp,0,sizeof(dp));
15         if(s==0)
16         {
17             printf("0.00000
");
18             continue;
19         }
20         dp[1][s] = 2;
21         dp[1][s-1] = 1;
22         for(i = 2; i <= n ;i++)
23             for(j = 0 ; j <= s ; j++)
24             {
25                 dp[i][j] = dp[i-1][j]*2+dp[i-1][j+1];
26             }
27         double ss=0;
28         for(i = 0; i <= s ; i++)
29         {
30             ss+=dp[n][i]/MM;
31         }
32         dp[n][0] = dp[n][0]/MM;
33         double x = 1-1.0*dp[n][0]/ss;
34         printf("%.5lf
",x*100);
35     }
36     return 0;
37 }
    View Code
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  • 原文地址：https://www.cnblogs.com/shangyu/p/3237752.html
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