Educational Codeforces Round 23D

给n个数求每个子区间的价值,区间的价值是最大值-最小值

套路题= =,分别算最大值和最小值的贡献,用并查集维护,把相邻点连一条边,然后sort,求最大是按边价值(两个点的最大价值)小的排,求最小是按最大排

类似的题:http://www.cnblogs.com/acjiumeng/p/8320666.html

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pii pair<int,int>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-9;
const int N=1000000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

struct edge{
    int u,v,ma,mi;
}e[N];
int father[N],sz[N],val[N];
int Find(int x)
{
    return father[x]==x?x:father[x]=Find(father[x]);
}
bool cmp1(edge a,edge b)
{
    return max(val[a.u],val[a.v])<max(val[b.u],val[b.v]);
}
bool cmp2(edge a,edge b)
{
    return min(val[a.u],val[a.v])>min(val[b.u],val[b.v]);
}
int main()
{
    int n,cnt=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&val[i]);
        if(i>1)
        {
            e[cnt].u=i-1,e[cnt].v=i;
            cnt++;
        }
        father[i]=i,sz[i]=1;
    }
    sort(e,e+cnt,cmp1);
//    for(int i=0;i<cnt;i++)
//        printf("%d %d
",e[i].u,e[i].v);
    ll ans=0;
    for(int i=0;i<cnt;i++)
    {
        int x=Find(e[i].u),y=Find(e[i].v);
        if(x!=y)
        {
            ans+=(ll)sz[x]*sz[y]*max(val[e[i].u],val[e[i].v]);
            father[x]=y,sz[y]+=sz[x];
        }
    }
    for(int i=1;i<=n;i++)father[i]=i,sz[i]=1;
    sort(e,e+cnt,cmp2);
    for(int i=0;i<cnt;i++)
    {
        int x=Find(e[i].u),y=Find(e[i].v);
        if(x!=y)
        {
            ans-=(ll)sz[x]*sz[y]*min(val[e[i].u],val[e[i].v]);
            father[x]=y,sz[y]+=sz[x];
        }
    }
    printf("%lld
",ans);
    return 0;
}
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