bzoj1609 [Usaco2008 Feb]Eating Together麻烦的聚餐（dp）

dp[i][1/2/3]表示把前i个变成以1/2/3结尾的合法队列的最少改动数，正着倒着各做一遍取最小。O(n)

#include <cstdio>#include <cstring>#define N 30005int n,a[N],dp[N][4],ans=0;inline int min(int x,int y){return x<y?x:y;}int main(){//  freopen("a.in","r",stdin);    scanf("%d",&n);    for(int i=1;i<=n;++i) scanf("%d",&a[i]);    memset(dp,0,sizeof(dp));    for(int i=1;i<=n;++i){        dp[i][1]=dp[i-1][1]+(a[i]==1?0:1);        dp[i][2]=min(dp[i-1][2],dp[i-1][1])+(a[i]==2?0:1);        dp[i][3]=min(min(dp[i-1][2],dp[i-1][1]),dp[i-1][3])+(a[i]==3?0:1);    }    ans=min(dp[n][1],min(dp[n][2],dp[n][3]));    for(int i=n;i>=1;--i){        dp[i][1]=dp[i+1][1]+(a[i]==1?0:1);        dp[i][2]=min(dp[i+1][2],dp[i+1][1])+(a[i]==2?0:1);        dp[i][3]=min(min(dp[i+1][2],dp[i+1][1]),dp[i+1][3])+(a[i]==3?0:1);    }    ans=min(min(ans,dp[1][1]),min(dp[1][2],dp[1][3]));    printf("%d\n",ans);    return 0;}