前言：嗯,这个题上午有的思路，敲了一中午代码，改了一下午最后超时?
题：D. Boris and His Amazing Haircut
题意：一个理发师可以把一段数组给建成一个高度，他现在每个高度的剪子都有若干个。给一个原始数组和目标数组，问能不能剪成目标数组。
奇怪的想法： 理发师最优的剪法就是先剪出来高的，然后在高的头发形成的段之间再剪次高的。直接上分治，结果超时，边界控制太难写了，
超时的代码：
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int length = 2e5 + 5;
vector<int> ag;
vector<int> bg;
map<int,int> mp;
vector<int> count1;
vector<vector<int>> index1;
int tree[length << 3];
void update(int l, int r, int L, int R, int cnt, int v)
{
if (L >= l && R <= r)
{
tree[cnt] = v;
return;
}
if (L > r || R < l)
{
return;
}
int mid = (L + R) / 2;
update(l, r, L, mid, cnt << 1, v);
update(l, r, mid + 1, R, cnt * 2 + 1, v);
tree[cnt] = max(tree[cnt * 2], tree[cnt * 2 + 1]);
}
int query(int l, int r, int L, int R, int cnt)
{
if (L>=l&&R<=r)
{
return tree[cnt];
}
if (L > r || R < l)
return 0;
int mid = (L + R) / 2;
int a = query(l, r, L, mid, cnt * 2);
int b = query(l, r, mid + 1, R, cnt * 2 + 1);
return max(a, b);
}
int n1;
bool DP(int l, int r)
{
//if()
if (l > r)
return 1;
int yh = query(l, r, 0, n1 - 1, 1);
bool sum = 1;
int lflag = ((l==0)?0:1);
int rflag = ((r==n1-1)?0:-1);
int cut = 0;
//找到第一个大于等于l的数
int a = -1;
int b = index1[yh].size();
while (a != b - 1)
{
int mid = (a + b) / 2;
if (index1[yh][mid] < l)
a = mid;
else
b = mid;
}
int i = b;
int in = 0;
for (; i<index1[yh].size()&&index1[yh][i]<=r; i++)
{
if (i != index1[yh].size() && mp[ag[index1[yh][i]]] != mp[bg[index1[yh][i]]]&&cut==0)
{
cut = 1;
if (count1[mp[bg[index1[yh][i]]]])
count1[mp[bg[index1[yh][i]]]]--;
else
return 0;
}
if (in == 0)
{
bool a=DP(l , index1[yh][i] - 1);
sum = sum & a;
}
else
{
bool a = DP(index1[yh][i - 1] + 1, index1[yh][i] - 1);
sum = sum & a;
}
if (sum == 0)
return sum;
in = 1;
}
if (i!=index1[yh].size()&&index1[yh][i] > r)
{
if (i != 0)
{
bool a = DP(index1[yh][i - 1] + 1, r);
sum = sum & a;
}
}
else if (i == index1[yh].size() && index1[yh][i - 1] != r)
{
bool a = DP(index1[yh][i - 1] + 1, r);
sum = sum & a;
}
return sum;
}
int main(void)
{
int t;
scanf_s("%d", &t);
for (int i = 0; i < t; i++)
{
int n;
scanf_s("%d", &n);
n1 = n;
vector<int> a(n,0);
for (int i = 0; i < n; i++)
{
scanf_s("%d", &a[i]);
}
vector<int> b(n, 0);
for (int i = 0; i < n; i++)
{
scanf_s("%d", &b[i]);
}
int m;
scanf_s("%d", &m);
vector<int> c(m, 0);
for (int i = 0; i < m; i++)
{
scanf_s("%d", &c[i]);
}
ag = a; bg = b;//全局变量赋值
vector<int> tmp=b;
sort(tmp.begin(), tmp.end());
vector<int>::iterator p = unique(tmp.begin(), tmp.end());
int off = p - tmp.begin();
map<int, int> hk;
int cnt = 1;
for (int i = 0; i < off; i++)
{
hk[tmp[i]] = i + 1;
}
mp = hk;
vector<int> tmp_count(length);
for (int i = 0; i < m; i++)
{
tmp_count[mp[c[i]]]++;//统计映射之后的数量
}
count1 = tmp_count;
memset(tree, 0, sizeof(tree));
vector<vector<int>> edge(length);
int ok = 1;
for (int i = 0; i < n; i++)
{
if (a[i] < b[i])
ok = 0;
update(i, i, 0, n - 1, 1, mp[b[i]]);//更新区间最值以及坐标位置
edge[mp[b[i]]].push_back(i);
}
if (ok == 0)
{
printf("NO\n");
continue;
}
index1 = edge;
int yh=DP(0, n - 1);
if (yh)
printf("YES\n");
else
printf("NO\n");
}
}
正解：
思考源头：为啥要剪一段？省剪子啊，他的剪子都是一次性的，用完不能再用乐。并且剪成高度为
5
5
5的时候不能穿过高度为
8
8
8的。所以就用单调栈呗，注意
a
[
i
]
=
=
b
[
i
]
a[i]==b[i]
a[i]==b[i]的头发可以弹出数，但是不能入栈，入栈的目的就是不让两边相同高度的头发剪两次。但是
a
[
i
]
=
=
b
[
i
]
a[i]==b[i]
a[i]==b[i]的可以作为山丘的两端，高度低的不能穿过他。
A
C
AC
AC代码：
#include<iostream>
#include<cstdio>
#include<map>
#include<stack>
using namespace std;
const int length = 2e5 + 5;
int a[length];
int b[length];
map<int, int> mp;
int main(void)
{
int t;
scanf_s("%d", &t);
for (int i = 0; i < t; i++)
{
mp.clear();
int n;
scanf_s("%d", &n);
for (int i = 0; i < n; i++)
{
scanf_s("%d", &a[i]);
}
for (int i = 0; i < n; i++)
{
scanf_s("%d", &b[i]);
}
int m;
scanf_s("%d", &m);
for (int i = 0; i < m; i++)
{
int a;
scanf_s("%d", &a);
mp[a]++;
}
int flag = 1;
stack<int> s;
for (int i = 0; i < n; i++)
{
if (a[i] < b[i])
{
flag = 0;
break;
}
while (!s.empty() && s.top() < b[i])
s.pop();
if ((s.empty() || s.top() > b[i])&&a[i]!=b[i])
{
s.push(b[i]);
if (mp[b[i]] == 0)
{
flag = 0;
break;
}
mp[b[i]]--;
}
}
if (flag == 0)
{
printf("NO\n");
}
else
printf("YES\n");
}
}
还有就是这个题的数值太大了，本来想用离散化的，但是如果能用 m a p map map的话直接用 m a p map map吧， n n n是够的。
