LuoguP1979 华容道 题解

宽搜

思路很简单，开一个四维数组used[i][j][k][l]表示指定点在(i, j),空格在(j, k)是否走过，每次让空格上下左右走，遇到指定旗子就改变旗子位置，时间复杂度(O(qn^4))

于是，我们得到了一份70分的代码：

#include <bits/stdc++.h>
using namespace std;
int mp[31][31];
struct hehe{
    int x, y, dx, dy, step;
};
bool used[31][31][31][31];
int mx[] = {0, 0, 1, 0, -1};
int my[] = {0, 1, 0, -1, 0};
int main()
{
    int n, m, T;
    cin >> n >> m >> T;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
            cin >> mp[i][j];
    }
    while(T--)
    {
        bool fl = 0;
        queue<hehe>q;
        memset(used, 0, sizeof(used));
        int qx, qy, zx, zy, kx, ky;
        cin >> qx >> qy >> zx >> zy >> kx >> ky;
        hehe now;
        now.dx = qx;
        now.dy = qy;
        now.x = zx;
        now.y = zy;
        now.step = 0;
        used[qx][qy][zx][zy] = 1;
        q.push(now);
        while(q.size())
        {
            now = q.front();
            q.pop();
            int x = now.x, y = now.y, xx = now.dx, yy = now.dy;
            if(x == kx && y == ky)
            {
                fl = 1;
                cout << now.step << endl;
                break;
            }
            for(int i = 1; i <= 4; i++)
            {
                int xxx = xx + mx[i], yyy = yy + my[i];
                if(xxx > n || yyy > m || xxx < 1 || yyy < 1) continue;
                int qwq = x, qaq = y;
                if(x == xxx && y == yyy) qwq = xx, qaq = yy;
                if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
                {
                    used[xxx][yyy][qwq][qaq] = 1;
                    hehe nxt;
                    nxt.x = qwq;
                    nxt.y = qaq;
                    nxt.dx = xxx;
                    nxt.dy = yyy;
                    nxt.step = now.step + 1;
                    q.push(nxt);
                }
            }
        }
        if(!fl)
        {
            cout << -1 << endl;
        }
    }
}

接下来才是这篇题解的重点：

如何卡常

O2优化+关闭流同步

#pragma GCC optimize(2)

上面的语句是开启O2优化

虽然在你谷不让在程序中自己写

但是可以手动开啊）

---

ios::sync_with_stdio(false);

这句话是关闭cin、cout的流同步

原理大概是让cin、cout执行过程中不另开缓存

会让cin、cout跑得比scanf、printf都快

---

加上这两句话，成功拿到了另外10分：

循环展开

定睛一看：嘶，#7#8被优化到了可观测的TLE之内（1.20s以内）

想想之前看过一个神仙卡常毒瘤题P4604 [WC2017]挑战

其中一个基本操作就是“循环展开”

具体一点：

以下代码

for(int i = 1; i <= 4; i++)
{
	int xxx = xx + mx[i], yyy = yy + my[i];
	if(xxx > n || yyy > m || xxx < 1 || yyy < 1) continue;
	int qwq = x, qaq = y;
	if(x == xxx && y == yyy) qwq = xx, qaq = yy;
	if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
	{
		used[xxx][yyy][qwq][qaq] = 1;
		hehe nxt;
		nxt.x = qwq;
		nxt.y = qaq;
		nxt.dx = xxx;
		nxt.dy = yyy;
		nxt.step = now.step + 1;
		q.push(nxt);
	}
}

将其循环展开后变成这样：

int i = 1;
int xxx = xx + mx[i], yyy = yy + my[i];
if(xxx > n || yyy > m || xxx < 1 || yyy < 1) goto s1;
qwq = x, qaq = y;
if(x == xxx && y == yyy) qwq = xx, qaq = yy;
if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
{
    used[xxx][yyy][qwq][qaq] = 1;
    hehe nxt;
    nxt.x = qwq;
    nxt.y = qaq;
    nxt.dx = xxx;
    nxt.dy = yyy;
    nxt.step = now.step + 1;
    q.push(nxt);
}
s1: i++;
xxx = xx + mx[i], yyy = yy + my[i];
if(xxx > n || yyy > m || xxx < 1 || yyy < 1) goto s2;
qwq = x, qaq = y;
if(x == xxx && y == yyy) qwq = xx, qaq = yy;
if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
{
    used[xxx][yyy][qwq][qaq] = 1;
    hehe nxt;
    nxt.x = qwq;
    nxt.y = qaq;
    nxt.dx = xxx;
    nxt.dy = yyy;
    nxt.step = now.step + 1;
    q.push(nxt);
}
s2: i++;
xxx = xx + mx[i], yyy = yy + my[i];
if(xxx > n || yyy > m || xxx < 1 || yyy < 1) goto s3;
qwq = x, qaq = y;
if(x == xxx && y == yyy) qwq = xx, qaq = yy;
if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
{
    used[xxx][yyy][qwq][qaq] = 1;
    hehe nxt;
    nxt.x = qwq;
    nxt.y = qaq;
    nxt.dx = xxx;
    nxt.dy = yyy;
    nxt.step = now.step + 1;
    q.push(nxt);
}
s3: i++;
xxx = xx + mx[i], yyy = yy + my[i];
if(xxx > n || yyy > m || xxx < 1 || yyy < 1) continue;
qwq = x, qaq = y;
if(x == xxx && y == yyy) qwq = xx, qaq = yy;
if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
{
    used[xxx][yyy][qwq][qaq] = 1;
    hehe nxt;
    nxt.x = qwq;
    nxt.y = qaq;
    nxt.dx = xxx;
    nxt.dy = yyy;
    nxt.step = now.step + 1;
    q.push(nxt);
}

这样可以优化常数——因为这个循环每次bfs新节点的时候都要做一次，这样理论上相当于优化了(10^8)级别的常数——虽然达不到

但是效果依然很明显：

好，#7#8已经在0.01级别了！#11#13也优化到可视范围内了！

优化掉i++和move数组

这两个微不足道的东西足以卡掉零点几秒的时间，甚至更多。

变成：

int xxx = xx + 1, yyy = yy + 0;
s1:
xxx = xx + 0, yyy = yy + 1;
...
s2:
xxx = xx - 1, yyy = yy + 0;
...
s3:
xxx = xx + 0, yyy = yy - 1;
...

好耶！多过了两个点！顺便把#11#13优化成了个位数的毫秒级！

最后，去掉goto

苦思冥想了好久，想起在goto身上还有一个常数

于是把判断去掉——因为在第0行、第0列、第n+1行、第m+1列都是零

将goto优化掉了

if(xxx > n || yyy > m || xxx < 1 || yyy < 1) goto s1;
...

去掉。

惊险AC。

最终代码

#include <bits/stdc++.h>
using namespace std;
int mp[31][31];
struct hehe{
    int x, y, dx, dy, step;
};
bool used[31][31][31][31];
int main()
{
    ios::sync_with_stdio(false);
    int n, m, T;
    cin >> n >> m >> T;
    for(register int i = 1; i <= n; i++)
    {
        for(register int j = 1; j <= m; j++)
            cin >> mp[i][j];
    }
    while(T--)
    {
        bool fl = 0;
        queue<hehe>q;
        memset(used, 0, sizeof(used));
        int qx, qy, zx, zy, kx, ky;
        cin >> qx >> qy >> zx >> zy >> kx >> ky;
        hehe now;
        now.dx = qx;
        now.dy = qy;
        now.x = zx;
        now.y = zy;
        now.step = 0;
        used[qx][qy][zx][zy] = 1;
        q.push(now);
        while(q.size())
        {
            now = q.front();
            q.pop();
            int x = now.x, y = now.y, xx = now.dx, yy = now.dy;
            if(x == kx && y == ky)
            {
                fl = 1;
                cout << now.step << endl;
                break;
            }
            int qwq, qaq;
                int xxx = xx + 1, yyy = yy + 0;
                if(xxx < n || yyy < m || xxx > 1 || yyy > 1)
                {
                qwq = x, qaq = y;
                if(x == xxx && y == yyy) qwq = xx, qaq = yy;
                if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
                {
                    used[xxx][yyy][qwq][qaq] = 1;
                    hehe nxt;
                    nxt.x = qwq;
                    nxt.y = qaq;
                    nxt.dx = xxx;
                    nxt.dy = yyy;
                    nxt.step = now.step + 1;
                    q.push(nxt);
                }
            }
                xxx = xx + 0, yyy = yy + 1;
                if(xxx < n || yyy < m || xxx > 1 || yyy > 1)
                {
                qwq = x, qaq = y;
                if(x == xxx && y == yyy) qwq = xx, qaq = yy;
                if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
                {
                    used[xxx][yyy][qwq][qaq] = 1;
                    hehe nxt;
                    nxt.x = qwq;
                    nxt.y = qaq;
                    nxt.dx = xxx;
                    nxt.dy = yyy;
                    nxt.step = now.step + 1;
                    q.push(nxt);
                }
            }
                xxx = xx - 1, yyy = yy + 0;
                if(xxx < n || yyy < m || xxx > 1 || yyy > 1)
                {
                    qwq = x, qaq = y;
                    if(x == xxx && y == yyy) qwq = xx, qaq = yy;
                    if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
                    {
                        used[xxx][yyy][qwq][qaq] = 1;
                        hehe nxt;
                        nxt.x = qwq;
                        nxt.y = qaq;
                        nxt.dx = xxx;
                        nxt.dy = yyy;
                        nxt.step = now.step + 1;
                        q.push(nxt);
                    }
                }
                xxx = xx + 0, yyy = yy - 1;
                if(xxx < n || yyy < m || xxx > 1 || yyy > 1)
                {
                qwq = x, qaq = y;
                if(x == xxx && y == yyy) qwq = xx, qaq = yy;
                if(mp[xxx][yyy] && !used[xxx][yyy][qwq][qaq])
                {
                    used[xxx][yyy][qwq][qaq] = 1;
                    hehe nxt;
                    nxt.x = qwq;
                    nxt.y = qaq;
                    nxt.dx = xxx;
                    nxt.dy = yyy;
                    nxt.step = now.step + 1;
                    q.push(nxt);
                }
                }
        }
        if(!fl)
        {
            cout << -1 << endl;
        }
    }
}

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