Solution CF1437E Make It Increasing
题目大意:给定一个序列,以及一些不能被修改的位置。问最少修改几个数可以使得序列严格单调递增。
dp
分析:
首先经典结论,把每个数减去它的下标,这样我们就只需要使得序列单调不降就可以了,元素可以相同更方便一些。
最少修改几个数,可以转化为,我们要在序列中选取一个最长不下降子序列,修改其他位置的值。而那些不能被修改的位置,就是我们必须选入子序列的位置。
这样我们沿用 (nlogn) 的经典做法,只不过稍加改动。设上一个钦定被选择的位置为 (pre),下一个位置为 (nxt),那么当前的位置 (p) 可能被选入子序列,当且仅当 (val[pre] leq val[p] leq val[nxt])
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#include <vector>
#pragma GCC optmize(2)
using namespace std;
typedef long long ll;
constexpr int maxn = 5e5 + 100,inf = 0x7fffffff;
struct IO{//-std=c++11,with cstdio and cctype
private:
static constexpr int ibufsiz = 1 << 20;
char ibuf[ibufsiz + 1],*inow = ibuf,*ied = ibuf;
static constexpr int obufsiz = 1 << 20;
char obuf[obufsiz + 1],*onow = obuf;
const char *oed = obuf + obufsiz;
public:
inline char getchar(){
#ifndef ONLINE_JUDGE
return ::getchar();
#else
if(inow == ied){
ied = ibuf + sizeof(char) * fread(ibuf,sizeof(char),ibufsiz,stdin);
*ied = ' ';
inow = ibuf;
}
return *inow++;
#endif
}
template<typename T>
inline void read(T &x){
static bool flg;flg = 0;
x = 0;char c = getchar();
while(!isdigit(c))flg = c == '-' ? 1 : flg,c = getchar();
while(isdigit(c))x = x * 10 + c - '0',c = getchar();
if(flg)x = -x;
}
template <typename T,typename ...Y>
inline void read(T &x,Y&... X){read(x);read(X...);}
inline int readi(){static int res;read(res);return res;}
inline long long readll(){static long long res;read(res);return res;}
inline void flush(){
fwrite(obuf,sizeof(char),onow - obuf,stdout);
fflush(stdout);
onow = obuf;
}
inline void putchar(char c){
#ifndef ONLINE_JUDGE
::putchar(c);
#else
*onow++ = c;
if(onow == oed){
fwrite(obuf,sizeof(char),obufsiz,stdout);
onow = obuf;
}
#endif
}
template <typename T>
inline void write(T x,char split = ' '){
static unsigned char buf[64];
if(x < 0)putchar('-'),x = -x;
int p = 0;
do{
buf[++p] = x % 10;
x /= 10;
}while(x);
for(int i = p;i >= 1;i--)putchar(buf[i] + '0');
if(split != ' ')putchar(split);
}
inline void lf(){putchar('
');}
~IO(){
fwrite(obuf,sizeof(char),onow - obuf,stdout);
}
}io;
template <typename A,typename B>
inline void chkmin(A &x,const B &y){if(y < x)x = y;}
template <typename A,typename B>
inline void chkmax(A &x,const B &y){if(y > x)x = y;}
int val[maxn],g[maxn],ans;
vector<int> pos;
int main(){
const int n = io.readi(),k = io.readi();
for(int i = 1;i <= n;i++)io.read(val[i]),val[i] -= i;
pos.push_back(0);
for(int i = 1;i <= k;i++)pos.push_back(io.readi());
for(unsigned int i = 2;i < pos.size();i++)
if(val[pos[i - 1]] > val[pos[i]])return io.write(-1,'
'),0;
pos.push_back(n + 1);
val[0] = -inf;
val[n + 1] = inf;
memset(g,0x7f,sizeof(g));
for(int i = 1;i <= n;i++){
int nxt = *upper_bound(pos.begin(),pos.end(),i);
int pre = *prev(lower_bound(pos.begin(),pos.end(),i));
if(val[i] > val[nxt] || val[i] < val[pre])continue;
int p = upper_bound(g + 1,g + n,val[i]) - g;
if(nxt == n + 1)chkmax(ans,p);
g[p] = val[i];
}
return io.write(n - ans,'
'),0;
}