CodeForces 935E Fafa and Ancient Mathematics (树形DP)

时间：2023-03-08 19:00:06

题意：给定一个表达式，然后让你添加 n 个加号，m 个减号，使得表达式的值最大。

析：首先先要建立一个表达式树，这个应该很好建立，就不说了，dp[u][i][0] 表示 u 这个部分表达式，添加 i 个符号，使值最大，dp[u][i][1] 表示 u 个部分表达式，添加 i 个符号，使用值最小，这里添加的符号可能是加号，也可以是减号，就是最小的那个，因为题目说了min(n, m) <= 100，一开始我没看到，就 WA8 了。然后在状转移的时候，分成两类，一类是添加的是加号，那么 dp[u][i][0] = max{dp[lson][j][0] + dp[rson][i-j][0], dp[lson][j][0] - dp[rson][i-j-1][1] }，同理可以得到添加减号的时候。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e4 + 20;

const int maxm = 1e6 + 10;

const LL mod = 1000000000000000LL;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int ch[maxn][2];

int num[maxn];

string s;

int rt, x;

void dfs(int l, int r, int &rt){

  if(l == r){

    rt = l;

    ms(ch[rt], -1);

    return ;

  }

  int det = 0;

  for(int i = l; i <= r; ++i)

    if(s[i] == '(')  ++det;

    else if(s[i] == ')')  --det;

    else if(s[i] == '?' && det == 1){

      rt = i;

      dfs(l+1, i-1, ch[i][0]);

      dfs(i+1, r-1, ch[i][1]);

      num[rt] = num[ch[i][0]] + num[ch[i][1]] + 1;

      return ;

    }

}

int dp[maxn][105][2];

void dfs(int u){

  if(ch[u][0] == -1){

    dp[u][0][0] = dp[u][0][1] = s[u] - '0';

    return ;

  }

  int tt = min(num[u], x);

  int ls = ch[u][0], rs = ch[u][1];

  int t = min(tt, num[ls]);

  dfs(ls);  dfs(rs);

  for(int i = 0; i <= tt; ++i){

    int &mmax = dp[u][i][0];  mmax = -INF;

    int &mmin = dp[u][i][1];  mmin = INF;

    for(int j = 0; j <= t; ++j){

      if(i - j > 0 && i - j - 1 <= min(x, num[rs]))  mmax = max(mmax, dp[ls][j][0] + (n <= m ? dp[rs][i-j-1][0] : -dp[rs][i-j-1][1]));

      if(i - j >= 0 && i - j <= min(num[rs], x))  mmax = max(mmax, dp[ls][j][0] - (n <= m ? dp[rs][i-j][1] : -dp[rs][i-j][0]));

      if(i - j > 0 && i - j - 1 <= min(x, num[rs]))  mmin = min(mmin, dp[ls][j][1] + (n <= m ? dp[rs][i-j-1][1] : -dp[rs][i-j-1][0]));

      if(i - j >= 0 && i - j <= min(num[rs], x))  mmin = min(mmin, dp[ls][j][1] - (n <= m ? dp[rs][i-j][0] : -dp[rs][i-j][1]));

    }

  }

}

int main(){

  cin >> s;

  cin >> n >> m;

  x = min(m, n);

  dfs(0, s.sz-1, rt);

  dfs(rt);

  cout << dp[rt][x][0] << endl;

  return 0;

}

　　

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