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=101000;
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=1;i<=n;i++) scanf("%lld%lld",&A[i].d,&A[i].p);
     
    sort(A+1,A+n+1,cmp); Cnt=0;
    for (LL i=1;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
",Ans);
    return 0;
}

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=2010;
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>0?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=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];
    for (int i=1;i<=n;i++) F[i][0]=Inf;
    sort(b+1,b+n+1,cmp1);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            C[i][j]=F[i-1][j]+Abs(b[j]-a[i]);
            F[i][j]=Min(F[i][j-1],C[i][j]);
        }
    Ans=F[n][n];
    sort(b+1,b+n+1,cmp2);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            C[i][j]=F[i-1][j]+Abs(b[j]-a[i]);
            F[i][j]=Min(F[i][j-1],C[i][j]);
        }
    printf("%d
",Min(Ans,F[n][n]));
    return 0;
}

 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=100500;
struct EDGE{LL to,next,w;}edge[Maxn<<1];
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!=-1;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!=-1;i=edge[i].next)
    {
        if (edge[i].to==fa) continue;
        if (Size[1]-2*Size[edge[i].to]<0)
        {
            Ans+=(Size[1]-2*Size[edge[i].to])*edge[i].w;
            Move(edge[i].to,u);
        }
    }
}
int main()
{
    scanf("%lld",&n);
    for (LL i=1;i<=n;i++) scanf("%lld",&C[i]);
    memset(head,-1,sizeof(head));
    for (LL i=1;i<n;i++) 
    {
        scanf("%lld%lld%lld",&u,&v,&w);
        Add(u,v,w),Add(v,u,w);
    }
    Ans=Dfs(1,0);
    Move(1,0);
    printf("%lld
",Ans);
    return 0;
}