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$jleq x_i$和$j>x_i$来做
$$ herefore f_{i,x_{i-1}} =
egin{cases}
f_{i-1,j}+x_i-jquad(jleq x_i)\
f_{i-1,j}+j-x_iquad(j>x_i)
end{cases}$$
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 ", V[1][0]); return 0; }