各种蕴含算法思想的DP - 1

study from:

https://www.cnblogs.com/flashhu/p/9480669.html

1.前缀和

https://www.luogu.org/problemnew/show/P2513

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <time.h>
 #include <string>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <bitset>
 #include <ext/rope>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long
 #define minv 1e-6
 #define inf 1e9
 #define pi 3.1415926536
 #define E  2.7182818284
 const int mod=1e4;//
 const int maxn=1e3+;
 
 int f[maxn][maxn],tot[maxn];
 
 int main()
 {
     int n,k,i,j,l;
     scanf("%d%d",&n,&k);
     f[][]=;
     for (i=;i<=n;i++)
     {
         tot[]=;
         l=min(i*(i-)/,k);
         for (j=;j<=l;j++)
             tot[j]=(tot[j-]+f[i-][j])%mod;
         for (j=;j<=l;j++)
             f[i][j]=(tot[j]-tot[max(j-i,-)]+mod)%mod;
     }
     printf("%d",f[n][k]);
     return ;
 }

https://www.luogu.org/problemnew/show/P2511

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <time.h>
 #include <string>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <bitset>
 #include <ext/rope>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long
 #define minv 1e-6
 #define inf 1e9
 #define pi 3.1415926536
 #define E  2.7182818284
 const int mod=;//
 const int maxn=5e4+;
 
 int a[maxn],t[maxn],tot[maxn],f[maxn][];
 
 int main()
 {
     int n,c,x,y,i,j,k,l,r,m,result=;
     scanf("%d%d",&n,&c);
     c++;
     for (i=;i<=n;i++)
         scanf("%d",&a[i]);
     l=; r=5e7;
     while (l<=r)
     {
         m=(l+r)>>;
         j=;
         k=;
         for (i=;i<=n;i++)
             if (a[i]>m)
                 break;
             else if (k+a[i]>m)
             {
                 j++;
                 k=a[i];
             }
             else
                 k+=a[i];
         if (i==n+ && j<=c)
             r=m-;
         else
             l=m+;
     }
     printf("%d ",l);
 
     t[]=;
     for (i=;i<=n;i++)
         t[i]=t[i-]+a[i];
 
     f[][]=;
     x=,y=;
     for (j=;j<=c;j++)
     {
         tot[-]=;
         for (i=;i<=n;i++)
             tot[i]=(tot[i-]+f[i][y])%mod;
         //f[i][j]  f[][j-1]
         k=;
         f[][x]=;
         for (i=;i<=n;i++)
         {
             //[k,i)
             while (t[i]-t[k]>l)
                 k++;
             f[i][x]=(tot[i-]-tot[k-])%mod;
         }
         result=(result+f[n][x])%mod;
         x=x^;
         y=y^;
     }
     printf("%d",(result+mod)%mod);
     return ;
 }
 /*
 5 1
 3 3 4 5 3
 
 5 3
 1 1 1 1 1
 */

2.单调队列优化

https://www.luogu.org/problemnew/show/P1886

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <time.h>
 #include <string>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <bitset>
 #include <ext/rope>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long
 #define minv 1e-6
 #define inf 1e9
 #define pi 3.1415926536
 #define E  2.7182818284
 const int mod=;//
 const int maxn=1e6+;
 
 /**
 n个数字a[1]-a[n],每个大小为m的区间
 最小值和最大值
 **/
 
 int a[maxn],x[maxn];    ///x:记录下标
 
 int main()
 {
     int n,m,i,head,tail;
     scanf("%d%d",&n,&m);
     for (i=;i<=n;i++)
         scanf("%d",&a[i]);
     ///min
     head=,tail=;
     ///以i作为末尾的区间
     for (i=;i<=n;i++)
     {
 
         while (head1<=tail1 && x[head1]<=i-m)
             head1++;
         while (head1<=tail1 && a[x[tail1]]>=a[i])
             tail1--;
 
         ///每个点入队列一次，最多出队列一次
         while (head<=tail && x[head]<=i-m)   ///在区间外的数字被剔除,(i-m,i]
             head++;
         while (head<=tail && a[x[tail]]>=a[i])   ///求区间最小/大值，对于在前面的数，只有数值小/大于后面才有用，否则从右往左被剔除(符号相反，大于等于/小于等于)，类似单调栈
             tail--;
         tail++;
         x[tail]=i;
         if (i>=m)
             printf("%d%c",a[x[head]],i==n?'\n':' ');
     }
 
     ///max
     head=,tail=;
     for (i=;i<=n;i++)
     {
         while (head<=tail && x[head]<=i-m)
             head++;
         while (head<=tail && a[x[tail]]<=a[i])
             tail--;
         tail++;
         x[tail]=i;
         if (i>=m)
             printf("%d%c",a[x[head]],i==n?'\n':' ');
     }
     return ;
 }