[AT2558]Many Moves

题目大意：有$n$个位置$1,2,\dots n$；你有两个棋子$A$和$B$，你要进行$q$次操作，第$i$次操作给定一个$x_i$，你要选择一个棋子移动到$x_i$；求两个棋子最小移动的步数之和。

题解：一个$O(n^2)$的$DP$容易想到$f_{i,j}$表示到了第$i$步，另一个棋子在$j$这个位置。

$$f_{i,x_{i-1}}=\min\{f_{i-1,j}+|x_i-j|\}$$

$$f_{i,j}=f_{i-1,j}+|x_i-x_{i-1}|$$

下面一个还好做，可上面一个呢？

可以考虑拆成$j\leq x_i$和$j>x_i$来做

$$\therefore f_{i,x_{i-1}} =
\begin{cases}
f_{i-1,j}+x_i-j\quad(j\leq x_i)\\
f_{i-1,j}+j-x_i\quad(j>x_i)
\end{cases}$$

然后发现是区间修改求最小值，可以用线段树来做。

卡点：1.转移时把$x_{i-1}$写成了$x_{i}$

C++ Code：

#include <cstdio>
#include <cstring>
#define maxn 200010
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
inline long long min(long long a, long long b) {return a < b ? a : b;}
inline long long abs(long long a) {return a > 0 ? a : -a;}
int n, k, x, y;
int q, last;
long long V[maxn << 2][3], cov[maxn << 2];
void pushdown(int rt) {int lc = rt << 1, rc = rt << 1 | 1;long long &tmp = cov[rt];V[lc][0] += tmp;V[lc][1] += tmp;V[lc][2] += tmp;cov[lc] += tmp;V[rc][0] += tmp;V[rc][1] += tmp;V[rc][2] += tmp;cov[rc] += tmp;tmp = 0;
}
void update(int rt) {int lc = rt << 1, rc = rt << 1 | 1;V[rt][0] = min(V[lc][0], V[rc][0]);V[rt][1] = min(V[lc][1], V[rc][1]);V[rt][2] = min(V[lc][2], V[rc][2]);
}
void add(int rt, int l, int r, int p, long long num) {if (l == r) {V[rt][0] = num;V[rt][1] = num + l;V[rt][2] = num - l;return ;}if (cov[rt]) pushdown(rt);int mid = l + r >> 1;if (p <= mid) add(rt << 1, l, mid, p, num);else add(rt << 1 | 1, mid + 1, r, p, num);update(rt);
}
void add(long long num, int rt = 1) {V[rt][0] += num;V[rt][1] += num;V[rt][2] += num;cov[rt] += num;
}
long long ask(int rt, int l, int r, int L, int R, int op) {if (L <= l && R >= r) return V[rt][op];pushdown(rt);int mid = l + r >> 1;long long ans = inf;if (L <= mid) ans = ask(rt << 1, l, mid, L, R, op);if (R > mid) ans = min(ans, ask(rt << 1 | 1, mid + 1, r, L, R, op));return ans;
}
int main() {scanf("%d%d%d%d", &n, &k, &x, &y);scanf("%d", &q);memset(V, 0x3f, sizeof V);add(1, 1, n, x, abs(y - q));add(1, 1, n, y, abs(x - q));while (--k) {last = q; scanf("%d", &q);long long t0 = ask(1, 1, n, last, last, 0) + abs(q - last);long long t1 = ask(1, 1, n, q, n, 1) - q;long long t2 = ask(1, 1, n, 1, q, 2) + q;long long ans = min(t0, min(t1, t2));add(abs(q - last));add(1, 1, n, last, ans);}printf("%lld\n", V[1][0]);return 0;
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9466135.html

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