[JXOI2018]守卫

嘟嘟嘟

正如某题解所说，这题很有误导性：我就一直在想凸包。

随便一个数据，就能把凸包hack掉：

这样我们的点G就gg了。

所以正解是什么呢？dp。

题解看这位老哥的吧，我感觉挺好懂的：题解 P4563 【[JXOI2018]守卫】

#include<cstdio>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<vector>

#include<stack>

#include<queue>

using namespace std;

#define enter puts("")

#define space putchar(' ')

#define Mem(a, x) memset(a, x, sizeof(a))

#define In inline

typedef long long ll;

typedef double db;

const int INF = 0x3f3f3f3f;

const db eps = 1e-8;

const int maxn = 5e3 + 5;

inline ll read()

{

  ll ans = 0;

  char ch = getchar(), last = ' ';

  while(!isdigit(ch)) last = ch, ch = getchar();

  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();

  if(last == '-') ans = -ans;

  return ans;

}

inline void write(ll x)

{

  if(x < 0) x = -x, putchar('-');

  if(x >= 10) write(x / 10);

  putchar(x % 10 + '0');

}

int n;

struct Point

{

  ll x, y;

  In Point operator + (const Point& oth)const

  {

    return (Point){x + oth.x, y + oth.y};

  }

  In Point operator - (const Point& oth)const

  {

    return (Point){x - oth.x, y - oth.y};

  }

  In ll operator * (const Point& oth)const

  {

    return x * oth.y - y * oth.x;

  }

}p[maxn];

int dp[maxn][maxn];

int main()

{

  n = read();

  for(int i = 1; i <= n; ++i) p[i].x = i, p[i].y = read();

  int ans = 0;

  for(int i = 1; i <= n; ++i)

    {

      dp[i][i] = 1; ans ^= 1;

      int sum = 1, pos = 0;

      for(int j = i - 1; j; --j)

	{

	  if(!pos || (p[pos] - p[i]) * (p[j] - p[i]) < 0)

	    sum += min(dp[j + 1][pos - 1], dp[j + 1][pos]), pos = j;

	  dp[j][i] = sum + min(dp[j][pos - 1], dp[j][pos]);

	  ans ^= dp[j][i];

	}

    }

  write(ans), enter;

  return 0;

}

巴特西

[JXOI2018]守卫

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