单调栈+RMQ
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5696
题意
给出一个长度为n(1 <= n <= 100000)的序列,区间权值为区间最大值乘以区间最小值,输出长度为1到n的区间的最大权值。
题解
先利用单调栈算出以a[i]为最小值的最大区间延伸,设该区间宽度为L,则更新ans[L](这里用到st表查询该区间最大值),最后从宽度较大的区间刷新宽度较小的区间的长度,即为答案。
为什么是对的,可以证明一下,从结果考虑,宽度为k的区间的最优值一定可以被这种方式取到。
ac代码:
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long lint;
#define maxn (100010)
#define left lleft
#define right rright
#define log2 Log2
int a[maxn], left[maxn], right[maxn];
lint ans[maxn];
int log2[maxn], st[maxn][32];
void st_prepare(int arr[], int n)
{
log2[1] = 0;
for(int i = 2; i <= n; i++)
{
log2[i] = log2[i-1];
if((1 << log2[i] + 1) == i)
log2[i]++;
}
for(int i = n-1; i >= 0; i--)
{
st[i][0] = arr[i];
for(int j = 1; i + (1 << j) - 1 < n; j++)
{
st[i][j] = max(st[i][j-1], st[i+(1<<j-1)][j-1]);
}
}
}
int st_query(int l, int r)
{
int len = log2[r - l + 1];
return max(st[l][len], st[r - (1 << len) + 1][len]);
}
#define stack Stack
pair<int, int> stack[maxn];
int top;
void push_stack(int id, int w, int val[], int mx)
{
while(top > 0)
{
if(stack[top].first < w)
break;
top--;
}
if(!top) val[id] = mx;
else val[id] = stack[top].second;
stack[++top] = make_pair(w, id);
}
int main()
{
int n;
while(cin >> n)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
st_prepare(a + 1, n);
top = 0;
for(int i = 1; i <= n; i++)
{
push_stack(i, a[i], left, 0);
}
top = 0;
for(int i = n; i >= 1; i--)
{
push_stack(i, a[i], right, n + 1);
}
for(int i = 0; i <= n; i++) ans[i] = -1;
for(int i = 1, l, r, k; i <= n; i++)
{
l = left[i];
r = right[i] - 2;
k = st_query(l, r);
ans[r - l + 1] = max(ans[r - l + 1], (lint)k * a[i]);
}
for(int i = n - 1; i >= 1; i--)
ans[i] = max(ans[i], ans[i + 1]);
for(int i = 1; i <= n; i++)
printf("%I64d\n", ans[i]);
}
return 0;
}
自己代码:(wa了,但没找到错误,先存起来,日后去看一下自己哪里错了)
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <stack>
using namespace std;
typedef long long ll;
const int N=100010;
ll arr[N];
ll dp[N][35],n;
ll L[N],R[N];
ll ans[N];
//rmq的建立
void rmq_init(){
for(int i=1;i<=n;i++)
dp[i][0]=arr[i];
for(int i=1;(1<<i)<=N;i++)
for(int j=1;j+(1<<i)<=N;j++)
dp[j][i]=max(dp[j][i-1],dp[j+(1<<i)-1][i-1]);
}
//rmq的查询
ll rmq_qui(int l,int r){
int k=(int)log(double(r-l+1))/log(2.0);
return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
void single_stack_l(){
stack<int >S;
while(S.size()) S.pop();
for(int i=1;i<=n;i++){
while(S.size()&&arr[S.top()]>=arr[i]) S.pop();
if(S.empty()) L[i]=0;
else L[i]=S.top();
S.push(i);
}
}
void single_stack_r(){
stack<int>S;
while(S.size()) S.pop();
for(int i=n;i>=1;i--){
while(S.size()&&arr[S.top()]>=arr[i]) S.pop();
if(S.empty()) R[i]=n+1;
else R[i]=S.top();
S.push(i);
}
}
int main(){
int a,b;
while(~scanf("%lld",&n)){
for(int i=1;i<=n;i++)
scanf("%lld",&arr[i]);
rmq_init();
int q,p;
// while(cin>>q>>p){
// cout<<rmq_qui(q,p)<<endl;
// }
single_stack_l();
single_stack_r();
//for(int i=1;i<=n;i++)
// printf("%d ",L[i]);
// printf("\n");
// for(int i=1;i<=n;i++)
// printf("%d ",R[i]);
//printf("\n");
for(int i=0;i<=n;i++) ans[i]=-1;
for(int i=1,l,r,k;i<=n;i++){
l=L[i];
r=R[i]-2;
k=rmq_qui(l+1,r+1);
ans[r-l+1]=max(ans[r-l+1],(ll)k*arr[i]);
//cout<<k<<" "<<ans[r-l+1]<<" "<<k*arr[i]<<endl;
}
for(int i=n-1;i>=1;i--)
ans[i]=max(ans[i],ans[i+1]);
for(int i=1;i<=n;i++)
printf("%lld\n",ans[i]);
}
return 0;
}