【Hihocoder1636】Pangu and Stones（区间DP）

题意：N堆石子，每次可以合并连续的长度从L到R的若干堆石子为1堆，费用为选择的石子总个数，求将N堆合并成1堆的最小总花费，无解输出0

思路：dp[i][j][k]表示将i到j这段区间合并为k堆的最小代价

\[ 初始条件 　　dp[i][j][j-i+1]=0 \]

\[ dp[i][j][k]=min(dp[i][x][y-1]+dp[x+1][j][1]+s[j]-s[i-1] 　　(k=1,i<=x<=j-1,L<=y<=R) \]

\[ dp[i][j][k]=min(dp[i][x][k-1]+dp[x+1][j][1]　　 (k>=2,i<=x<=j-1) \]

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   150
 #define M   6100000
 #define eps 1e-8
 #define pi  acos(-1)
 #define oo  1e9
 
 ll dp[N][N][N],a[N],s[N];
 
 int main()
 {
     //freopen("hihocoder1636.in","r",stdin);
     //freopen("hihocoder1636.out","w",stdout);
     int n,L,R;
     while(scanf("%d%d%d",&n,&L,&R)!=EOF)
     {
         s[]=;
         for(int i=;i<=n;i++)
         {
             scanf("%lld",&a[i]);
             s[i]=s[i-]+a[i];
         }
         memset(dp,0x3f,sizeof(dp));
         for(int i=;i<=n;i++)
          for(int j=i;j<=n;j++) dp[i][j][j-i+]=;
         for(int len=;len<=n;len++)
          for(int i=;i<=n-len+;i++)
          {
              int j=i+len-;
              for(int x=i;x<=j-;x++)
               for(int y=L;y<=R;y++)
               dp[i][j][]=min(dp[i][j][],dp[i][x][y-]+dp[x+][j][]+s[j]-s[i-]);
              for(int k=;k<=len;k++)
               for(int x=i;x<=j-;x++)
                dp[i][j][k]=min(dp[i][j][k],dp[i][x][k-]+dp[x+][j][]);
          }
         if(dp[][n][]>oo) printf("0\n");
          else printf("%lld\n",dp[][n][]);
     }
     return ;
 }