单调栈简单题目整理
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2022-03-20 18:51:21
最大矩阵类问题1.求最大矩阵题目链接题目:直方图中的最大矩阵#include#define pb push_back#define fi first#define se second#define ALL(a) a.begin(), a.end()#define SZ(a) (int)a.size()#define db1(x) cout<<#x<<" = "<
最大矩阵类问题
1.求最大矩阵
题目:直方图中的最大矩阵
#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
//#define int long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6+10;
int n;
ll h[N];
int w[N], p[N];
void sol() {
for(int i = 1;i <= n; ++i) scanf("%lld", &h[i]);
h[n+1] = 0;
int c = 0;
ll ma = 0;
for(int i = 1;i <= n+1; ++i) {
ll x = h[i];
if(h[i]>=p[c]) {
p[++c] = h[i];
w[c] = 1;
}
else {
int width = 0;
while(h[i]<p[c]) {
width += w[c];
ma = max(ma, 1ll*width*p[c--]);
}
p[++c] = h[i]; w[c] = width+1;
}
}
printf("%lld\n", ma);
}
int main()
{
while(scanf("%d", &n), n)
sol();
return 0;
}
2 求最大正方形,也就是最大矩阵稍微改改
题目:城市游戏
需要先预处理从当前位置向右边连续的F的个数O(n^2)
然后枚举起始列,对每一列进行一个单调栈处理出最大正方形O(n^2)
#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<string, int> pii;
typedef pair<int, int> piii;
typedef unsigned long long ull;
const int N = 2010, maxn = 1e6+10;
const ll mod = 1e9+7;
int n, m, k;
char s[N][N];
int num[N][N];
int calc(int x)
{
int h[N], w[N], p[N], c = 0;
for(int i = 1; i <= n; ++i) h[i] = num[i][x], w[i] = p[i] = 0;
h[0] = 0;
h[n+1] = 0;
p[0] = 0;
int ans = 0;
for(int i = 1;i <= n+1; ++i) {
if(h[i] >= p[c]) p[++c] = h[i], w[c] = 1;
else {
int width = 0;
while(h[i] < p[c]) {
width += w[c];
ans = max(ans, width*p[c--]);
}
p[++c] = h[i], w[c] = width+1;
}
}
return ans;
}
void sol() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j) scanf(" %c", &s[i][j]);
for(int i = 1;i <= n; ++i)
for(int j = 1;j <= m; ++j)
{
if(s[i][j] == 'F') num[i][j] = 1;
else continue;
if(num[i][j-1]>1&&s[i][j]==s[i][j-1]) {
num[i][j] = num[i][j-1]-1;
}
else {
int k = j;
while(k+1 <= m&&s[i][j]==s[i][k+1]) k++, num[i][j]++;
}
}
int ans = 0;
for(int st = 1; st <= m; ++st)
ans = max(ans, calc(st));
printf("%lld\n", 3ll*ans);
}
signed main()
{
sol();
}
最近的最高或最低的位置问题。
1 模板类
题目:仰视奶牛
#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<string, int> pii;
typedef pair<int, int> piii;
typedef unsigned long long ull;
const int N = 2e5+10, maxn = 1e6+10;
const ll mod = 1e9+7;
int n, m, k;
int h[N], p[N], ans[N];
void sol() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &h[i]);
int c = 0;
for(int i = 1; i <= n; ++i)
{
if(c==0||h[i]<=h[p[c]]) p[++c] = i;
else {
while(c!=0&&h[i]>h[p[c]]) {
ans[p[c--]] = i;
}
p[++c] = i;
}
}
for(int i = 1;i <= n; ++i)
printf("%d\n", ans[i]);
}
signed main()
{
sol();
}
2 接雨水类
题目:接雨水
维护一个单调严格递减的栈,若当前柱子高于栈顶的柱子,退栈直到符合单调性;否则加入栈顶。每次退栈时,累加雨水的体积(每次算一层)
#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef unsigned long long ull;
const int N = 2e5+10, maxn = 1e6+10;
const ll mod = 1e9+7;
int n, m, k;
int h[N];
void sol() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &h[i]);
stack<int> st;
ll ans = 0;
for(int i = 1;i <= n; ++i)
{
while(!st.empty()&&h[st.top()]<h[i]) {
int t = st.top(); st.pop();
if(!st.empty()){
int l = i-st.top()-1;
int w = min(h[i], h[st.top()])-h[t];
ans += l*w;
}
}
st.push(i);
}
printf("%lld\n", ans);
}
signed main()
{
sol();
}
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