UVa 1625 Color Length (DP)

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;
}