USACO 刷水

BZOJ 1666 水..

BZOJ 1579 分层图最短路.

BZOJ 1782 从一开始若某头牛停在U，那么U的子树的时间都会加一用BIT维护DFS序就行了

BZOJ 1572

贪心+堆 排序后查看是否超过了时间，超过了就弹出

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #define LL long long
 using namespace std;
 const LL Maxn=;
 struct Node{LL p,d;}A[Maxn];
 LL Ans,Cnt,n;
 priority_queue<LL> Q;
 inline bool cmp(Node A,Node B)
 {return A.d<B.d || (A.d==B.d && A.p<B.p);}
 int main()
 {
     scanf("%lld",&n);
     for (LL i=;i<=n;i++) scanf("%lld%lld",&A[i].d,&A[i].p);

     sort(A+,A+n+,cmp); Cnt=;
     for (LL i=;i<=n;i++)
     {
         Ans+=A[i].p; Cnt++; Q.push(-A[i].p);
         if (Cnt>A[i].d) {Ans+=Q.top(); Q.pop(); Cnt--; }
     }
     printf("%lld\n",Ans);
     return ;
 }

BZOJ 1572

BZOJ 1592

这道题有个结论就是最小花费的高度肯定和在原来的高度中.

这个结论感觉上非常显然，如果不在原来的高度中，那个肯定能从原来的高度中找到最优的.然后直接Dp就可以了

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int Maxn=;
 const int Inf=0x3f3f3f3f;
 int a[Maxn],b[Maxn],C[Maxn][Maxn],F[Maxn][Maxn],Ans,n;
 inline int Min(int x,int y) {return x>y?y:x;}
 inline int Abs(int x) {return x>?x:-x;}
 inline bool cmp1(int a,int b) {return a<b;}
 inline bool cmp2(int a,int b) {return a>b;}
 int main()
 {
     scanf("%d",&n);
     for (int i=;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];
     for (int i=;i<=n;i++) F[i][]=Inf;
     sort(b+,b+n+,cmp1);
     for (int i=;i<=n;i++)
         for (int j=;j<=n;j++)
         {
             C[i][j]=F[i-][j]+Abs(b[j]-a[i]);
             F[i][j]=Min(F[i][j-],C[i][j]);
         }
     Ans=F[n][n];
     sort(b+,b+n+,cmp2);
     for (int i=;i<=n;i++)
         for (int j=;j<=n;j++)
         {
             C[i][j]=F[i-][j]+Abs(b[j]-a[i]);
             F[i][j]=Min(F[i][j-],C[i][j]);
         }
     printf("%d\n",Min(Ans,F[n][n]));
     return ;
 }

BZOJ 1592

BZOJ 1827

树形DP，U节点向子节点转移时，

ans′=ans–size∗w+(tot–size)∗w

ans′=ans+(tot–2∗size)∗w 若(tot–2∗size)<0那么ans'比ans优

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 #define LL long long
 using namespace std;

 const LL Maxn=;
 struct EDGE{LL to,next,w;}edge[Maxn<<];
 LL head[Maxn],C[Maxn],n,u,v,w,Ans,Dis[Maxn],Size[Maxn],cnt;
 inline void Add(LL u,LL v,LL w)
 {edge[cnt].to=v;edge[cnt].next=head[u];edge[cnt].w=w;head[u]=cnt++;}
 LL Dfs(LL u,LL fa)
 {
     LL Ret=Dis[u]*C[u]; Size[u]=C[u];
     for (LL i=head[u];i!=-;i=edge[i].next)
     {
         if (edge[i].to==fa) continue;
         Dis[edge[i].to]=Dis[u]+edge[i].w;
         Ret+=Dfs(edge[i].to,u);
         Size[u]+=Size[edge[i].to];
     }
     return Ret;
 }
 void Move(LL u,LL fa)
 {
     for (LL i=head[u];i!=-;i=edge[i].next)
     {
         if (edge[i].to==fa) continue;
         if (Size[]-*Size[edge[i].to]<)
         {
             Ans+=(Size[]-*Size[edge[i].to])*edge[i].w;
             Move(edge[i].to,u);
         }
     }
 }
 int main()
 {
     scanf("%lld",&n);
     for (LL i=;i<=n;i++) scanf("%lld",&C[i]);
     memset(head,-,sizeof(head));
     for (LL i=;i<n;i++)
     {
         scanf("%lld%lld%lld",&u,&v,&w);
         Add(u,v,w),Add(v,u,w);
     }
     Ans=Dfs(,);
     Move(,);
     printf("%lld\n",Ans);
     return ;
 }

BZOJ 1827