当时我的第一想法也是用单调栈,但是被我写炸了;我也不知道错在哪里;
看了大神的写法,用数组模拟的;
记录下单调递增栈的下标,以及每个数字作为最小值的最左边的位置。
当有数据要出栈的时候,说明栈里的数据已经不是最小了,右端点就是当前位置-1,那么就可以计算栈顶的元素所作的贡献;出栈完后,当前这个数字,他的最左边就是栈顶所能到达的位置;入栈;
#include <bits/stdc++.h> using namespace std; const int maxn = + ;
int a[maxn];
int stacks[maxn];
long long sum[maxn];
int lef[maxn]; int main()
{
freopen("feelgood.in","r",stdin);
freopen("feelgood.out","w",stdout);
int n;
scanf("%d",&n); memset(sum,,sizeof(sum));
memset(lef,,sizeof(lef)); for(int i=;i<=n;i++) {
scanf("%d",&a[i]);
sum[i] = sum[i-] + a[i];
} a[++n] = -;
int top = ;
long long ans = -;
int ansl = ,ansr = ;
for(int i=;i<=n;i++) {
if(top==||a[i]>a[stacks[top-]]) {
stacks[top++] = i;
lef[i] = i;
continue;
}
if(a[i]==a[stacks[top-]])
continue;
while(top>=&&a[i]<a[stacks[top-]]) {
top --;
long long tmp = (long long)a[stacks[top]]*(sum[i-]-sum[lef[stacks[top]]-]);
if(tmp>ans) {
ansr = i-;
ansl = lef[stacks[top]];
ans = tmp;
}
} lef[i] = lef[stacks[top]];
stacks[top++] = i; } printf("%lld\n%d %d\n",ans,ansl,ansr); return ;
}
(之前的错误找到了ans=-1,可以都为0)
#include <bits/stdc++.h> using namespace std; int n;
const int maxn = +;
struct num {
long long value;
int maxleft,maxright;
int minleft,minright;
num():maxleft(),maxright(),minleft(),minright(){}
}a[maxn]; stack<pair<int,int> > S; long long sum[maxn]; void getMax()
{
while(!S.empty())
S.pop();
S.push(make_pair(a[].value,));
for(int i=;i<n;i++) {
while(!S.empty()&&S.top().first<=a[i].value) {
//int value = S.top().first;
int key = S.top().second;
S.pop(); a[i].maxleft +=a[key].maxleft;
if(!S.empty()) {
a[S.top().second].maxright +=a[key].maxright;
}
}
S.push(make_pair(a[i].value,i));
}
while(!S.empty()) {
int key = S.top().second;
S.pop();
if(!S.empty()) {
a[S.top().second].maxright +=a[key].maxright;
}
}
} void getMin()
{
while(!S.empty())
S.pop();
S.push(make_pair(a[].value,));
for(int i=;i<n;i++) {
while(!S.empty()&&S.top().first>=a[i].value) {
//int value = S.top().first;
int key = S.top().second;
S.pop(); a[i].minleft +=a[key].minleft;
if(!S.empty()) {
a[S.top().second].minright +=a[key].minright;
}
}
S.push(make_pair(a[i].value,i));
}
while(!S.empty()) {
int key = S.top().second;
S.pop();
if(!S.empty()) {
a[S.top().second].minright +=a[key].minright;
}
}
} int main()
{
freopen("feelgood.in","r",stdin);
freopen("feelgood.out","w",stdout);
scanf("%d",&n);
for(int i=;i<n;i++) {
scanf("%lld",&a[i].value);
sum[i+] = sum[i] + a[i].value;
}
// getMax();
getMin(); int l = ;
int r = ;
long long ans = -;
for(int i=;i<n;i++) {
long long tmp = a[i].value*(sum[i+a[i].minright]-sum[i-a[i].minleft+]);
if(ans<tmp) {
ans = tmp;
l = i - a[i].minleft + ;
r = i + a[i].minright;
}
} printf("%lld\n%d %d\n",ans,l,r); return ;
}