题目链接
这道题不能直接用dp[][]等方式来寄存数据，所以得想办法去进行优化，于是乎，就是想到了线段树的优化作用，我们可以这样子去推一个状态转移，如图：
最后拍照下来的图像是反的…… QAQ……
但就是这样的一个意思，我们可以这样去推状态方程，然后记录在线段树上面即可。
具体可以参考一下代码。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 2e5 + 7;
#define MIN_3(a, b) min(INF, min(a, b))
int N, M;
char mp[3][maxN];
struct node
{
int dp[4];
node() { memset(dp, INF, sizeof(dp)); }
friend node operator + (node e1, node e2)
{
node ans;
ans.dp[0] = MIN_3(e1.dp[0] + e2.dp[0] + 1, e1.dp[1] + e2.dp[2] + 1);
ans.dp[1] = MIN_3(e1.dp[1] + e2.dp[3] + 1, e1.dp[0] + e2.dp[1] + 1);
ans.dp[2] = MIN_3(e1.dp[2] + e2.dp[0] + 1, e1.dp[3] + e2.dp[2] + 1);
ans.dp[3] = MIN_3(e1.dp[2] + e2.dp[1] + 1, e1.dp[3] + e2.dp[3] + 1);
return ans;
}
}tree[maxN<<2];
inline void pushup(int rt) { tree[rt] = tree[lsn] + tree[rsn]; }
void buildTree(int rt, int l, int r)
{
if(l == r)
{
if(mp[1][l] == '.') tree[rt].dp[0] = 0;
if(mp[2][l] == '.') tree[rt].dp[3] = 0;
if(mp[1][l] == '.' && mp[2][l] == '.') tree[rt].dp[1] = tree[rt].dp[2] = 1;
return;
}
int mid = HalF;
buildTree(Lson);
buildTree(Rson);
pushup(rt);
}
node query(int rt, int l, int r, int ql, int qr)
{
if(ql <= l && qr >= r) return tree[rt];
int mid = HalF;
if(qr <= mid) return query(QL);
else if(ql > mid) return query(QR);
else return query(QL) + query(QR);
}
int output(int x, int y)
{
int u = x > N ? (x - N) : x, v = y > N ? (y - N) : y;
if(u > v) { swap(u, v); swap(x, y); }
node ans = query(1, 1, N, u, v);
if(x <= N && y <= N) return ans.dp[0];
else if(x > N && y <= N) return ans.dp[2];
else if(x <= N && y > N) return ans.dp[1];
else return ans.dp[3];
}
int main()
{
scanf("%d%d", &N, &M);
for(int i=1; i<=2; i++) scanf("%s", mp[i] + 1);
buildTree(1, 1, N);
for(int i=1, x, y; i<=M; i++)
{
scanf("%d%d", &x, &y);
int ans = output(x, y);
printf("%d\n", ans < INF ? ans : -1);
}
return 0;
}