dp bzoj 2436 noi2011 noi嘉年华

https://www.lydsy.com/JudgeOnline/problem.php?id=2436

只会\(\mathcal O(n^5)\)dp怎么办

首先必要的离散化以及预处理一下区间\([i,j]\)的线段数量\(num[i][j]\)

注意线段是左闭右开的

记\(f[i][j]\)为在\([1, i]\)中A拿了\(j\)个B最多拿\(f[i][j]\)个（状态设定的时候

转移枚举\(k\)
\[ f[i][j] = max\left \{ f[k][j] + num[k][i],f[k][j-num[k][i]] \right \} \]
同理\(g[i][j]\)为在\([i,\infty)\)中A拿了\(j\)个B最多拿\(g[i][j]\)个

转移类似

此部分复杂度\(\mathcal O(n^3)\)

记\(dp[i][j]\)为强制A选\([i,j]\)的最优答案

转移将\(f,g,num\)三部分合并在一起

枚举A在\(f,g\)分别选了\(x,y\)

感性理解可知随着\(x\)的单增\(y\)不减

因此复杂度\(\mathcal O(n^3)\)

总复杂度也是\(\mathcal O(n^3)\)

在状态比较维数比较多的时候可以考虑利用增加辅助的\(dp\)数组来降低复杂度

#include <bits/stdc++.h>
#define fo(i, n) for(int i = 1; i <= (n); i ++)
#define out(x) cerr << #x << " = " << x << "\n"
using namespace std;
// by piano
template<typename tp> inline void read(tp &x) {
  x = 0; char c = getchar(); bool f = 0;
  for(; c < '0' || c > '9'; f |= (c == '-'), c = getchar());
  for(; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - '0', c = getchar());
  if(f) x = -x;
}
const int N = 1111;
struct NODE {
  int l, r, id;
}p[N];
int f[N][N], g[N][N], num[N][N], dp[N][N], pcnt[N][N], ans[N];
int n, x[N], cnt = 0;
inline bool chkmax(int &x, int y) {
  return x < y ? x = y, 1 : 0;
}
inline bool cmp(NODE a, NODE b) {
  return (a.l < b.l) || (a.l == b.l && a.r < b.r);
}
#define cal(i, j, x, y)                                 \
  (max(min(x + y + num[i][j], f[i][x] + g[j][y]),       \
      min(x + y, f[i][x] + g[j][y] + num[i][j])))

main(void) {
  read(n);
  for(int i = 1; i <= n; i ++) {
    read(p[i].l); read(p[i].r); p[i].r += p[i].l;
    x[++ cnt] = p[i].l; x[++ cnt]= p[i].r; p[i].id = i;
  }
  sort(p + 1, p + n + 1, cmp);
  sort(x + 1, x + cnt + 1);
  cnt = unique(x + 1, x + cnt + 1) - x - 1;
  
  for(int i = 1; i <= n; i ++) {
    p[i].l = lower_bound(x + 1, x + cnt + 1, p[i].l) - x;
    p[i].r = lower_bound(x + 1, x + cnt + 1, p[i].r) - x;
    ++ pcnt[p[i].l][p[i].r];
  }

  memset(num, 0, sizeof num);
  for(int len = 1; len <= cnt; len ++) {
    for(int l = 1; l <= cnt; l ++) {
      int r = l + len - 1; if(r > cnt) break;
      num[l][r] = pcnt[l][r] + num[l + 1][r] + num[l][r - 1] - num[l + 1][r - 1];
    }
  }
  
  for(int i = 1; i <= cnt; i ++)
    for(int j = 1; j <= n; j ++)
      f[i][j] = g[i][j] = -6666;
  
  f[0][0] = 0;
  for(int i = 1; i <= cnt; i ++) {
    for(int j = 0; j <= num[1][i]; j ++) {
      chkmax(f[i][j], f[i - 1][j]);
      for(int k = 1; k < i; k ++) {
        chkmax(f[i][j], f[k][j] + num[k][i]);
        if(j - num[k][i] >= 0)
          chkmax(f[i][j], f[k][j - num[k][i]]); 
      }
    }
  }
  
  for(int j = 0; j <= n; j ++)
    chkmax(ans[0], min(j, f[cnt][j]));

  g[cnt + 1][0] = 0;
  for(int i = cnt; i >= 1; i --) {
    for(int j = 0; j <= num[i][cnt]; j ++) {
      chkmax(g[i][j], g[i + 1][j]);
      for(int k = cnt; k > i; k --) {
        chkmax(g[i][j], g[k][j] + num[i][k]);
        if(j - num[k][i] >= 0)
          chkmax(g[i][j], g[k][j - num[i][k]]); 
      }
    }
  }
  
  for(int i = 1; i <= cnt; i ++) {
    for(int j = i + 1; j <= cnt; j ++) {
      for(int x = 0, y = n; x <= n && y >= 0; x ++) {
          while(y) {
            int sb = cal(i, j, x, y - 1);
            int sn = cal(i, j, x, y);
            if(sb >= sn)
              -- y;
            else break;
          }
          chkmax(dp[i][j], cal(i, j, x, y));
        }
    }
  }

  for(int i = 1; i <= n; i ++)
    for(int l = 1; l <= p[i].l; l ++)
      for(int r = p[i].r; r <= cnt; r ++) {
        chkmax(ans[p[i].id], dp[l][r]);
  }

  for(int i = 0; i <= n; i ++)
    cout << ans[i] << "\n";
}

dp bzoj 2436 noi2011 noi嘉年华

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