【分治计数|单调栈】51Nod 1215 数组的宽度

题面在这里

用单调栈分别维护max和min的”势力范围”

显然很好做

其实分治的话就更简单了

只需要记录mx和mn以及它们的前缀和就好了

分类讨论一下，这是分治计数的核心

示例程序：

分治计数：

#include<cstdio>#include<algorithm>using namespace std;typedef long long LL;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}const int maxn=50005,INF=0x3f3f3f3f;int n,a[maxn],mx[maxn],mn[maxn];LL ans,smx[maxn],smn[maxn];void divide(int L,int R){    if (L==R) return;    int mid=L+R>>1;    divide(L,mid);divide(mid+1,R);    mx[mid]=0;mn[mid]=INF;    smx[mid]=smn[mid]=0;    for (int i=mid+1;i<=R;i++){        mx[i]=max(mx[i-1],a[i]);mn[i]=min(mn[i-1],a[i]);        smx[i]=smx[i-1]+mx[i];smn[i]=smn[i-1]+mn[i];    }    LL Max=0,Min=INF;    for (int i=mid,jx=mid,jm=mid;i>=L;i--){        Max=max(Max,(LL)a[i]);Min=min(Min,(LL)a[i]);        while (jx<R&&mx[jx+1]<=Max) jx++;        while (jm<R&&mn[jm+1]>=Min) jm++;        ans+=Max*(jx-mid)+smx[R]-smx[jx]-Min*(jm-mid)-smn[R]+smn[jm];    }}int main(){                  n=red();    for (int i=1;i<=n;i++) a[i]=red();    divide(1,n);    printf("%lld",ans);    return 0;}

单调栈：

#include<cstdio>#include<algorithm>#define LL long longusing namespace std;const int maxn=50005;int n,a[maxn],stk[maxn],len;LL l[maxn];int main(){    scanf("%d",&n);    for (int i=1;i<=n;i++) scanf("%d",&a[i]);    len=0;LL ans=0;    for (int i=1;i<=n;i++){        while (len&&a[stk[len]]<=a[i]) ans+=l[stk[len]]*(i-stk[len])*a[stk[len]],len--;        l[i]=i-stk[len];stk[++len]=i;    }    while (len) ans+=l[stk[len]]*(n-stk[len]+1)*a[stk[len]],len--;    for (int i=1;i<=n;i++){        while (len&&a[stk[len]]>=a[i]) ans-=l[stk[len]]*(i-stk[len])*a[stk[len]],len--;        l[i]=i-stk[len];stk[++len]=i;    }    while (len) ans-=l[stk[len]]*(n-stk[len]+1)*a[stk[len]],len--;    printf("%lld",ans);    return 0;}