HDU 3410 Passing the Message

可以先处理出每个a[i]最左和最右能到达的位置，L[i]，和R[i]。然后就只要询问区间[ L[i],i-1 ]和区间[ i+1,R[i] ]最大值位置即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
inline int read()
{
    char c = getchar();  while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

const int maxn=50000+10;
int T,n,a[maxn],L[maxn],R[maxn],dp[maxn][30];

void RMQ_init()
{
    for(int i=0;i<n;i++) dp[i][0]=i;
    for(int j=1;(1<<j)<=n;j++)
        for(int i=0;i+(1<<j)-1<n;i++)
        {
            if(a[dp[i][j-1]]>a[dp[i+(1<<(j-1))][j-1]]) dp[i][j]=dp[i][j-1];
            else dp[i][j]=dp[i+(1<<(j-1))][j-1];
        }
}

int RMQ(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=R-L+1) k++;
    if(a[dp[L][k]]>a[dp[R-(1<<k)+1][k]]) return dp[L][k];
    return dp[R-(1<<k)+1][k];
}

int main()
{
    scanf("%d",&T); int cas=1;
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",&a[i]), L[i]=R[i]=i;
        for(int i=1;i<n;i++)
        {
            if(a[i]<a[i-1]) continue; int pre=L[i-1];
            while(1) { L[i]=pre; if(pre==0||a[pre-1]>a[i]) break; pre=L[pre-1]; }
        }
        for(int i=n;i>=1;i--) R[i]=i;
        for(int i=n-2;i>=0;i--)
        {
            if(a[i]<a[i+1]) continue; int pre=R[i+1];
            while(1) { R[i]=pre; if(pre==n-1||a[pre+1]>a[i]) break; pre=R[pre+1]; }
        }

        RMQ_init();
        printf("Case %d:
",cas++);
        for(int i=0;i<n;i++)
        {
            if(L[i]>i-1) printf("0 "); else printf("%d ",RMQ(L[i],i-1)+1);
            if(R[i]<i+1) printf("0
"); else printf("%d
",RMQ(i+1,R[i])+1);
        }
    }
    return 0;
}