UVA 12170 Easy Climb(dp+单调队列优化)

题意：给出一个序列，首尾不可改变，要求你使序列满足相邻2个|x_i - x_(i-1)| <=d，改变的代价就是两数相减绝对值，问最小代价。

做法：通过枚举可以发觉每个数字只能变成a_i+k*d这种形式的数字，那么这样每个数的状态只有n^2个了，不过转移的话代价比较高，因为对于d(i,x)需要枚举d(i-1,y)| x - d <= y <= x+d最小的转移过来，其实这个就是相当于一个求区间最值的问题，我们可以利用单调队列做到线性的，下标递增，值也是递增的，如果当前值要大于队首，那么直接加上，反之就要把那些大于当前值的元素出队列，因为我当前要求的区间右边界已经到这里了，意味着最小值至少是当前这个了，比他大的自然就没用了。这是左边界的处理，右边界的话，就看是否满足x-d<=y即可。不满足就让队尾的出队，直至满足为止。

AC代码：

//#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<queue>#include<set>#include<map>#include<cmath>#include<ctime>#include<string.h>#include<string>#include<sstream>#include<bitset>using namespace std;#define ll long long#define ull unsigned long long#define eps 1e-8#define MOD 1000000007#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define PI acos(-1)#define NMAX (ll)1<<50template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}template<class T> inline T Max(T a, T b){ return a > b ? a : b; }template<class T> inline T Min(T a, T b){ return a < b ? a : b; }ll a[105],b[20000+200];ll dp[105][20000+200];ll que[20000+200][2];int nct;int getpos(int x){    return lower_bound(b,b+nct,x)-b;}ll cal(ll a, ll b){    ll ret = a-b;    if(ret < 0) ret = -ret;    return ret;}int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o1.txt","w",stdout);#endif // GLQ    int cas,n;    ll d;    scanf("%d",&cas);    while(cas--)    {        scanf("%d%lld",&n,&d);        for(int i = 0; i < n; i++)            scanf("%lld",&a[i]);        nct = 0;        for(int i = 0; i < n; i++)        {            b[nct++] = a[i];            for(int j = 1; j < n; j++)            {                b[nct++] = a[i]+(ll)j*d;                if(a[i]-(ll)j*d >= 0) b[nct++] = a[i]-(ll)j*d;            }        }        sort(b,b+nct);        nct = unique(b,b+nct)-b;        for(int i = 0; i < n; i++)            for(int j = 0; j < nct; j++)                dp[i][j] = NMAX;        dp[0][getpos(a[0])] = 0;        int front,rear;        for(int i = 0; i < n-1; i++)        {            front = -1; rear = 0;            int k = 0;            for(int j = 0; j < nct; j++)            {                while(k < nct && cal(b[k],b[j]) <= d)                {                    if(front >= rear && dp[i][k] <= que[front][1])                    {                        while(front >= rear && dp[i][k] <= que[front][1])                            front--;                    }                    que[++front][0] = k;                    que[front][1] = dp[i][k];                    k++;                }                while(cal(b[que[rear][0]],b[j]) > d) rear++;                dp[i+1][j] = Min(dp[i+1][j],que[rear][1]+cal(a[i+1],b[j]));            }        }        if(dp[n-1][getpos(a[n-1])] == NMAX) printf("impossible\n");        else printf("%lld\n",dp[n-1][getpos(a[n-1])]);    }    return 0;}