UVa 1625 Color Length (DP)

时间：2022-10-10 04:48:48

题意：给定两个序列，让你组成一个新的序列，让两个相同字符的位置最大差之和最小。组成方式只能从一个序列前部拿出一个字符放到新序列中。

析：这个题状态表示和转移很容易想到，主要是在处理上面，dp[i][j] 表示从第一序列中拿了 i 个字符，从第二序列中拿了 j 个字符的最小和是多少，这个要提前预处理每个字符开始出现和最后出现的位置，然后再用一个c数组来记录已经有多少个字符出现，但没有结束。注意要清空。

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>

#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 = 5000 + 10;

const int maxm = maxn * 100;

const int mod = 10;

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], c[maxn][maxn];

int f1[30], f2[30], r1[30], r2[30];

char s1[maxn], s2[maxn];

int a[maxn], b[maxn];

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%s", s1);

    scanf("%s", s2);

    n = strlen(s1);

    m = strlen(s2);

    memset(f1, INF, sizeof f1);

    memset(f2, INF, sizeof f2);

    memset(r1, 0, sizeof r1);  //must clear

    memset(r2, 0, sizeof r2);  //must clear

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

      a[i] = s1[i-1] - 'A';

      f1[a[i]] = min(f1[a[i]], i);

      r1[a[i]] = i;

    }

    for(int i = 1; i <= m; ++i){

      b[i] = s2[i-1] - 'A';

      f2[b[i]] = min(f2[b[i]], i);

      r2[b[i]] = i;

    }

    for(int i = 0; i <= n; ++i)

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

        if(!i && !j)  continue;

        int v1 = INF, v2 = INF;

        if(i)  v1 = dp[i-1][j] + c[i-1][j];

        if(j)  v2 = dp[i][j-1] + c[i][j-1];

        dp[i][j] = min(v1, v2);

        if(j){

          c[i][j] = c[i][j-1];

          if(f2[b[j]] == j && f1[b[j]] > i)  ++c[i][j];  // take care of this '>' not '>='

          if(r2[b[j]] == j && r1[b[j]] <= i)  --c[i][j];  //take care, too

        }

        else{

          c[i][j] = c[i-1][j];

          if(f1[a[i]] == i && f2[a[i]] > j)  ++c[i][j];

          if(r1[a[i]] == i && r2[a[i]] <= j)  --c[i][j];

        }

      }

    printf("%d\n", dp[n][m]);

  }

  return 0;

}

　　

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