[dp]HDOJ4960 Another OCD Patient

题意: 给一个n, 第二行给n堆的价值v[i], 第三行给a[i].  a[i]表示把i堆合在一起需要的花费.

　　求把n堆变成类似回文的 需要的最小花费.

思路:

①记忆化搜索 比较好理解...

dp[l][r] 记录l到r的最小花费

枚举对称轴 维护每次l到r之间对称

dp[l][r]=min(dp[l][r], a[cur-l]+a[r-i]+dfs(cur+1, i-1));

l左边和r右边的合并

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cmath>
 #include <string>
 #include <sstream>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 #include <queue>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <set>
 #include <map>
 typedef long long LL;
 typedef long double LD;
 #define pi acos(-1.0)
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 typedef pair<int, int> PI;
 typedef pair<int, PI> PP;
 #ifdef _WIN32
 #define LLD "%I64d"
 #else
 #define LLD "%lld"
 #endif
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
 inline void print(LL x){printf(LLD, x);puts("");}
 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}

 int a[];
 LL num[];
 int cost[];
 int dp[][];
 int dfs(int l, int r)
 {
     if(dp[l][r]!=-)
         return dp[l][r];
     if(l>=r)
         return dp[l][r]=;
     dp[l][r]=cost[r-l];
     int cur=l;
     for(int i=r;i>=l;i--)
     {
         for(;cur<i && (num[cur]-num[l-]<num[r]-num[i-]);cur++);
         if(cur==i)
             break;
         if(num[cur]-num[l-]==num[r]-num[i-])
             dp[l][r]=min(dp[l][r], cost[cur-l]+cost[r-i]+dfs(cur+, i-));
     }
     return dp[l][r];
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif
     int n;
     while(~scanf("%d", &n) && n)
     {
         num[]=;
         for(int i=;i<=n;i++)
         {
             scanf("%d", &a[i]);
             num[i]=num[i-]+(LL)a[i];
         }
         for(int i=;i<n;i++)
             scanf("%d", &cost[i]);
         for(int i=;i<=n;i++)
             fill(dp[i]+i+, dp[i]+n+, -);
         //memset(dp, -1, sizeof(dp));
         printf("%d\n", dfs(, n));
     }
     return ;
 }

HDOJ 4960 记忆化搜索

② 区间dp

结构体里有 SUM和NUM记录 左或右 NUM堆合起来 合成总值为SUM的

左边从1开始往右 右边从n开始往左

Lsum==Rsum了就分别把Lsum合起来 Rsum合起来 分别存进L和R   然后继续往中间

Ltmp和Rtmp分别记录左边和右边分别合到哪里了

L R 搞完以后 按顺序把 L 中间 R 放进final (注意中间)

因为对称 所以只要 dp  final 的一半就够了

dp[i]=min(dp[i], dp[j]+cost[左个数]+cost[右个数]);

最后要注意合了 i堆之后 把剩下的全合起来的情况

ans=min(ans, dp[i]+cost[剩下堆数]);

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cmath>
 #include <string>
 #include <sstream>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 #include <queue>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <set>
 #include <map>
 typedef long long LL;
 typedef long double LD;
 #define pi acos(-1.0)
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 typedef pair<int, int> PI;
 typedef pair<int, PI> PP;
 #ifdef _WIN32
 #define LLD "%I64d"
 #else
 #define LLD "%lld"
 #endif
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
 inline void print(LL x){printf(LLD, x);puts("");}
 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}

 struct node
 {
     LL SUM;
     int NUM;
 }L[], R[], final[];
 LL dp[];
 int a[];
 int cost[];
 int Ltop, Rtop, top;
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif
     int n;
     while(~scanf("%d", &n) && n)
     {
         memset(L, , sizeof(L));
         memset(R, , sizeof(R));
         memset(final, , sizeof(final));
         memset(a, , sizeof(a));
         memset(cost, , sizeof(cost));
         for(int i=;i<=n;i++)
             scanf("%d", &a[i]);
         for(int i=;i<=n;i++)
             scanf("%d", &cost[i]);

         LL Lsum=a[], Rsum=a[n];
         int Lnum=, Rnum=;
         Ltop=Rtop=;
         int Ltmp=, Rtmp=n;
         for(int i=, j=n; i<j;)
         {
             while(Lsum!=Rsum)
             {
                 if(i==j)
                     break;
                 if(Lsum<Rsum)
                 {
                     Lnum++;
                     Lsum+=a[++i];
                 }
                 else if(Rsum<Lsum)
                 {
                     Rnum++;
                     Rsum+=a[--j];
                 }
             }
             if(Lsum==Rsum)
             {
                 L[Ltop].SUM=Lsum, L[Ltop++].NUM=Lnum;
                 R[Rtop].SUM=Rsum, R[Rtop++].NUM=Rnum;
                 Lsum=a[++i];
                 Rsum=a[--j];
                 Lnum=Rnum=;
                 Ltmp=i;
                 Rtmp=j;
             }
         }
         top=;
         for(int i=;i<Ltop;i++)
             final[++top]=L[i];
         if(Ltmp<=Rtmp)
         {
             int sum=;
             for(int i=Ltmp;i<=Rtmp;i++)
                 sum+=a[i];
             final[++top].SUM=sum, final[top].NUM=Rtmp-Ltmp+;
         }
         for(int i=Rtop-;i>=;i--)
             final[++top]=R[i];
         for(int i=;i<=top/;i++)
         {
             int tmp1=final[i].NUM, tmp2=final[top-i+].NUM;
             dp[i]=LLONG_MAX;
             for(int j=i-;j>=;j--)
             {
                 dp[i]=min(dp[i], dp[j]+cost[tmp1]+cost[tmp2]);
                 tmp1+=final[j].NUM;
                 tmp2+=final[top-j+].NUM;
             }
         }
         LL minn=LLONG_MAX;
         LL ans=n;
         for(int i=;i<=top/;i++)
         {
             minn=min(minn, dp[i]+cost[ans]);
             ans-=final[i+].NUM+final[top-i].NUM;
         }
         print(minn);
     }
     return ;
 }

HDOJ 4960 区间dp