POJ 3280 Cheapest Palindrome (区间DP)

Posted 2020-09-05 dwtfukgv

题意：字串S长M，由N个小写字母构成。欲通过增删字母将其变为回文串，增删特定字母花费不同，求最小花费。

析：是一个简单DP，dp[i][j] 表示区间 i - j 是回文串的最小花费，很容易知道，删除和添加效果是一样的，所以我们就可以只取一个最小值就好。

做的时候我的初始化在外面，就一直WA。

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>
#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 freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -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;
}
int dp[maxn][maxn];
char s[maxn];
map<char, int> mp;

int main(){
  while(scanf("%d %d", &m, &n) == 2){
    scanf("%s", s);
    char op[5];
    for(int i = 0; i < m; ++i){
      int x, y;
      scanf("%s %d %d", op, &x, &y);
      mp[op[0]] = min(x, y);
    }
    memset(dp, 0, sizeof dp);
    for(int l = 1; l < n; ++l){
      for(int i = 0; i + l < n; ++i){
        int j = i + l;
        dp[i][j] = INF;
        if(s[i] == s[j]){  dp[i][j] = dp[i+1][j-1];  continue; }
        dp[i][j] = min(dp[i][j], dp[i+1][j] + mp[s[i]]);
        dp[i][j] = min(dp[i][j], dp[i][j-1] + mp[s[j]]);
      }
    }
    printf("%d\n", dp[0][n-1]);
  }
  return 0;
}