SDUT_经典线型DP

1001  即hdu 1003 Max Sum

状态转移方程 dp[i]=max(dp[i],dp[i-1]);没这么写

是相邻最大和的题，以前做过的题，数组开小了，wa了4次，最后才发现，弱暴了

#include<stdio.h>
int num[100005];
int main()
{

    int T,cas=1;
    scanf("%d",&T);
    while(cas<=T)
    {
        if(cas!=1) printf("\n");
        int n; scanf("%d",&n);
        for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);
        int sum=0,max=-99999,s=1,e=1,pos=1;
        for(int i=1;i<=n;i++)
        {

            if(sum<0)
            {sum=num[i];pos=i;}
            else sum+=num[i];

            if(max<sum)
            {
                s=pos;
                max=sum;
                e=i;
            }
        }
        printf("Case %d:\n%d %d %d\n",cas++,max,s,e);
    }
    return 0;
}

1002   即hdu 1081 To The Max

1001 这道题的变形，变成2维的了，自己写了个还是wa，还是没学好啊，百度下发现，思考不全面

先是1--n行枚举+一维的dp

#include<stdio.h>
#include<string.h>
int num[105][105],ss[105];
int main()
{
   int n;
   while(~scanf("%d",&n))
   {
       for(int i=0;i<n;i++)
         for(int j=0;j<n;j++)
         {
             scanf("%d",&num[i][j]);
         }
       int max=-99999;
       for(int i=0;i<n;i++)
       {
         memset(ss,0,sizeof(ss));
         for(int j=i;j<n;j++)
         {
             int t=0;
             for(int k=0;k<n;k++)
             {
                 ss[k]+=num[j][k];
                 if(t<=0) t=ss[k];
                 else t+=ss[k];
                 if(t>max) max=t;
             }
         }
       }
       printf("%d\n",max);
   }
   return 0;
}

1003   hdu 1025 Constructing Roads In JGShining's Kingdom

不知怎么的总是有tle的思路代码，这题呢，经典的lis 题，

#include<stdio.h>
#define N 500005
int map[N],flag[N];
int main()
{
    int n,cas=1;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            map[a]=b;
        }
        int low,hig,mid,top=1;
        flag[1]=map[1];
        for(int i=2;i<=n;i++)
        {
            low=1;hig=top;
            while(low<=hig)
            {
                mid=(low+hig)/2;
                if(flag[mid]<map[i])
                low=mid+1;
                else hig=mid-1;
            }
            flag[low]=map[i];
            if(low>top) top++;
        }
        printf("Case %d:\n",cas++);
        if(top==1)
             printf("My king, at most %d road can be built.\n\n",top);
        else
             printf("My king, at most %d roads can be built.\n\n",top);
    }
    return 0;
}

1004  即hdu1058 Humble Numbers

做过的题了，自己做的思路 感觉定位在 枚举还是比较合适，不合主题~o_o

2 3 5 7  和 前 n-1 个数从大到小枚举

#include<stdio.h>
__int64 str[5845];
int main()
{
    int n;
    int b[4]={2,3,5,7};
    str[1]=1;
    for(int i=2;i<=5842;i++)
    {
        str[i]=2000000005;
        for(int j=0;j<4;j++)
        {
          for(int k=i-1;k>=1;k--)
          {
              if(str[k]*b[j]<=str[i-1]) break;
              if(str[i]>str[k]*b[j])
              str[i]=str[k]*b[j];
          }
        }
    }
    while(scanf("%d",&n),n)
    {
        printf("The %d",n);
        if(n%10==1&&n%100!=11) printf("st");
        else if(n%10==2&&n%100!=12) printf("nd");
        else if(n%10==3&&n%100!=13) printf("rd");
        else printf("th");
        printf(" humble number is %I64d.\n",str[n]);
    }
    return 0;
}

1005  即hdu 1080 Human Gene Functions

自己没想出来思路，百度的

先把字符处理成int ，然后就是 dp转移方程 

grah[i][j]=max(grah[i-1][j-1]+score[ps[a[i-1]]][ps[b[j-1]]]
           ,max(grah[i][j-1]+score[4][ps[b[j-1]]],grah[i-1][j]+score[4][ps[a[i-1]]]));

经典啊，我的想法走歪了

#include<iostream>
#include<map>
#include<string>
#include<algorithm>
using namespace std;
map<char,int > ps;

int main()
{
    ps['A']=0;ps['C']=1;ps['G']=2;ps['T']=3;ps['-']=4;
    const int score[5][5]=
    {{5,-1,-2,-1,-3},
    {-1,5,-3,-2,-4},
    {-2,-3,5,-2,-2},
    {-1,-2,-2,5,-1},
    {-3,-4,-2,-1,0}};
    int n;cin>>n;
    while(n--)
    {
     string a,b;int lena,lenb;
     cin>>lena>>a>>lenb>>b;
     int grah[105][105];
     grah[0][0]=0;
     for(int i=1;i<=lena;i++)
     grah[i][0]=grah[i-1][0]+score[ps[a[i-1]]][4];
     for(int i=1;i<=lenb;i++)
     grah[0][i]=grah[0][i-1]+score[4][ps[b[i-1]]];

     for(int i=1;i<=lena;i++)
       for(int j=1;j<=lenb;j++)
       {
           grah[i][j]=max(grah[i-1][j-1]+score[ps[a[i-1]]][ps[b[j-1]]]
           ,max(grah[i][j-1]+score[4][ps[b[j-1]]],grah[i-1][j]+score[4][ps[a[i-1]]]));
       }
       cout<<grah[lena][lenb]<<endl;
    }
    return 0;
}