CodeForces 935E Fafa and Ancient Mathematics (树形DP)
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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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