51Nod-1572-宝岛地图

ACM模版

描述

题解

这个问题一开始我选择暴力，结果 TLE 了，然后各种优化，考虑到很多位置是重复判断的，所以我选择将所有可能会路过的相对坐标存在 set 中，结果依然是 TLE 了两组，虽然时间上快了很多，但是依然无法避免的超时了（代码 One）。

然后，发现自己虽然在不断地优化，但是始终没有摆脱暴力的思路，这里应该将有障碍的地方全部设置为 1，然后求前缀和，因为如果我们一步一步的去判断肯定会超时，我们应该一个指令一个指令去判断，每一个指令都是一个矩形的位移，所以我们可以根据前缀和来实现一下子判断（代码 Two）~~~

我竟然一开始没有想到这种优化，脑子被驴踢了，一个三级题难为我那么久……

代码

One：

/*
//  TLE
#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>

using namespace std;

const int MAXN = 1010;
const int MAX_LETTER = 26;
const int MAX_ORDER = 1e5 + 10;

int n, m, k;
char map[MAXN][MAXN];

struct node
{
int x, y;
} L[MAX_LETTER];

struct order
{
char dir;
int steps;
} O[MAX_ORDER];

set<pair<int, int> > spii;
set<pair<int, int> >::iterator it;

void charge(int x, int y)
{
for (int i = 0; i < k; i++)
{
for (int j = 0; j < O[i].steps; j++)
{
if (O[i].dir == 'N')
{
x -= 1;
}
else if (O[i].dir == 'S')
{
x += 1;
}
else if (O[i].dir == 'E')
{
y += 1;
}
else if (O[i].dir == 'W')
{
y -= 1;
}
spii.insert(make_pair(x, y));
}
}
}

int charge_(int x, int y)
{
int a, b;
for (it = spii.begin(); it != spii.end(); it++)
{
a = x + it->first;
b = y + it->second;
if (a < 0 || a >= n || y < 0 || y >= m || map[a][b] == '#')
{
return 0;
}
}
return 1;
}

int main(int argc, const char * argv[])
{
memset(L, -1, sizeof(L));

cin >> n >> m;
for (int i = 0; i < n; i++)
{
scanf("%s", map[i]);
for (int j = 0; j < m; j++)
{
if (map[i][j] >= 'A' && map[i][j] <= 'Z')
{
int k = map[i][j] - 'A';
L[k].x = i;
L[k].y = j;
}
}
}

cin >> k;
char dir[3];
for (int i = 0; i < k; i++)
{
scanf("%s%d", dir, &O[i].steps);
O[i].dir = dir[0];
}

int x = 0, y = 0;
spii.insert(make_pair(0, 0));
charge(x, y);

int flag = 0;
for (int i = 0; i < MAX_LETTER; i++)
{
if (L[i].x != -1)
{
if (charge_(L[i].x, L[i].y))
{
cout << (char)(i + 'A');
flag = 1;
}
}
}
if (flag == 0)
{
cout << "no solution";
}
puts("");

return 0;
}
*/

Two：

#include <algorithm>
#include <iostream>
#include <cstdio>

using namespace std;

const int MAXN = 1010;
const int MAX_LETTER = 26;
const int MAX_ORDER = 1e5 + 10;

struct node
{
int x, y;
} L[MAX_LETTER];

struct order
{
char dir;
int steps;
} O[MAX_ORDER];

int n, m, k;
char s[MAXN];
int map[MAXN][MAXN];

bool ok(int x, int y)
{
return x > 0 && y > 0 && x <= n && y <= m;
}

int Sum(int a, int b, int c, int d)
{
return map[a - 1][b - 1] + map[c][d] - map[a - 1][d] - map[c][b - 1];
}

bool charge(int x, int y)
{
for (int i = 1; i <= k; i++)
{
if (O[i].dir == 'N')
{
if (!ok(x - O[i].steps, y))
{
return 0;
}
if (Sum(x - O[i].steps, y, x, y))
{
return 0;
}
x = x - O[i].steps;
}
if (O[i].dir == 'S')
{
if (!ok(x + O[i].steps, y))
{
return 0;
}
if (Sum(x, y, x + O[i].steps, y))
{
return 0;
}
x = x + O[i].steps;
}
if (O[i].dir == 'W')
{
if (!ok(x, y - O[i].steps))
{
return 0;
}
if (Sum(x, y - O[i].steps, x, y))
{
return 0;
}
y = y - O[i].steps;
}
if (O[i].dir == 'E')
{
if (!ok(x, y + O[i].steps))
{
return 0;
}
if (Sum(x, y, x, y + O[i].steps))
{
return 0;
}
y = y + O[i].steps;
}
}
return 1;
}

int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
scanf("%s", s + 1);
for (int j = 1; j <= m; j++)
{
if (s[j] == '#')
{
map[i][j] = 1;
}
else if (s[j] >= 'A' && s[j] <= 'Z')
{
int k = s[j] - 'A';
L[k].x = i;
L[k].y = j;
}
}
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
map[i][j] = map[i][j] + map[i - 1][j] + map[i][j - 1] - map[i - 1][j - 1];
}
}

cin >> k;
for (int i = 1; i <= k; i++)
{
scanf("%s%d", s, &O[i].steps);
O[i].dir = s[0];
}

int flag = 0;
for (int i = 0; i < MAX_LETTER; i++)
{
if (L[i].x && charge(L[i].x, L[i].y))
{
putchar(i + 'A');
flag = 1;
}
}
if (!flag)
{
cout << "no solution";
}
putchar(10);

return 0;
}