UVA - 1625 Color Length[序列DP 提前计算代价]

if(bg[c][0]==i&&bg[c][1]>j) w[i][j]++;
if(ed[c][0]==i&&ed[c][1]<=j) w[i][j]--;

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int N=5005,INF=1e9;
int T,n,m;
char a
,b
;
int f

,w

,bg[27][2],ed[27][2];
void dp(){
memset(bg,0,sizeof(bg));
memset(ed,0,sizeof(ed));
for(int i=0;i<26;i++) bg[i][0]=bg[i][1]=INF;
for(int i=1;i<=n;i++){
int c=a[i]-'A';
if(bg[c][0]==INF) bg[c][0]=i;
ed[c][0]=i;
}
for(int i=1;i<=m;i++){
int c=b[i]-'A';
if(bg[c][1]==INF) bg[c][1]=i;
ed[c][1]=i;
}
for(int i=0;i<=n;i++)
for(int j=0;j<=m;j++){
//f[i][j]=min(f[i-1][j]+cal(i-1,j),f[i][j-1]+cal(i,j-1));
//f[i][j]=min(f[i-1][j],f[i][j-1])+cal(i,j);
if(i==0&&j==0) continue;
int t1=INF,t2=INF;
if(i>0) t1=f[i-1][j]+w[i-1][j];
if(j>0) t2=f[i][j-1]+w[i][j-1];
f[i][j]=min(t1,t2);

if(i>0){
w[i][j]=w[i-1][j];
int c=a[i]-'A';
if(bg[c][0]==i&&bg[c][1]>j) w[i][j]++;
if(ed[c][0]==i&&ed[c][1]<=j) w[i][j]--;
}else{
w[i][j]=w[i][j-1];
int c=b[j]-'A';
if(bg[c][1]==j&&bg[c][0]>i) w[i][j]++;
if(ed[c][1]==j&&ed[c][0]<=i) w[i][j]--;
//if(b[j]=='R') printf("R %d %d %d %d\n",bg[c][1],ed[c][1],i,j);
}
//printf("%d %d %d %d\n",i,j,f[i][j],w[i][j]);
}
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%s%s",a+1,b+1);
n=strlen(a+1);m=strlen(b+1);
dp();
printf("%d\n",f
[m]);
}
}

// UVa1625 Color Length
// Rujia Liu
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 5000 + 5;
const int INF = 1000000000;

char p[maxn], q[maxn]; // starts from position 1
int sp[26], sq[26], ep[26], eq[26]; // sp[i] start positions of character i in p
int d[2][maxn], c[2][maxn]; // 滚动数组 c[i][j]: how many "incomplete" colors in the mixed sequence

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%s%s", p+1, q+1);

int n = strlen(p+1);
int m = strlen(q+1);
for(int i = 1; i <= n; i++) p[i] -= 'A';
for(int i = 1; i <= m; i++) q[i] -= 'A';

// calculate s and e
for(int i = 0; i < 26; i++)
{
sp[i] = sq[i] = INF;
ep[i] = eq[i] = 0;
}
for(int i = 1; i <= n; i++)
{
sp[p[i]] = min(sp[p[i]], i);
ep[p[i]] = i;
}
for(int i = 1; i <= m; i++)
{
sq[q[i]] = min(sq[q[i]], i);
eq[q[i]] = i;
}

// dp
int t = 0;
memset(c, 0, sizeof(c));
memset(d, 0, sizeof(d));
for(int i = 0; i <= n; i++)
{
for(int j = 0; j <= m; j++)
{
if(!i && !j) continue;

// calculate d
int v1 = INF, v2 = INF;
//计算d[i][j],  d[i][j]由d[i-1][j]或d[i][j-1]添加一个字母得到
if(i) v1 = d[t^1][j] + c[t^1][j]; // remove from p
if(j) v2 = d[t][j - 1] + c[t][j - 1]; // remove from q
d[t][j] = min(v1, v2);

// calculate c
if(i)
{
c[t][j] = c[t^1][j];
if(sp[p[i]] == i && sq[p[i]] > j) c[t][j]++;            //出现新的字母
if(ep[p[i]] == i && eq[p[i]] <= j) c[t][j]--;           //一个字母已经结束
}
else if(j)
{
c[t][j] = c[t][j - 1];
if(sq[q[j]] == j && sp[q[j]] > i) c[t][j]++;
if(eq[q[j]] == j && ep[q[j]] <= i) c[t][j]--;
}
}
t ^= 1;
}
printf("%d\n", d[t^1][m]);
}
return 0;
}