版权声明：那个，最起码帮我加点人气吧，署个名总行吧 https://blog.csdn.net/qq_41670466/article/details/84701601
题意就是在一个m*n的地图上，有k个堡垒，每个堡垒都会想一个固定的方向射箭，现在问你能否在给定的体力下从（0，0）走到（m，n），
分析：其实这题的难点在于如何判定你当前走到的点是否有箭，一开始我想的是预处理，在输入每一个堡垒时，就对它的箭的路线进行标记，但是后来发现这样很难对两个箭相遇时处理，于是就换成了当你走到某一格时，去四个方向找堡垒看看它是否有射向那个点的键以及能否射中。
这个代码一开始总是莫名其妙的wa了，后来加了几个else break 就神奇的过了，哭;
代码：
#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
struct Node
{
int x, y;
int enr;
bool operator <(const Node b)const
{
return enr > b.enr;
}
}node[1010];
struct Cas
{
int x, y;
int t, v;
char di;
}cas[110];
bool vis[110][110][1010];
int map[110][110];
int dir[5][2] = { 1,0,-1,0,0,1,0,-1,0,0 };
int m, n, k, d;
int geth(Node b)//利用了一点a*算法的思想，对其进行预估
{
return abs(b.x - m) + abs(b.y - n);
}
int check(Node b)
{
if (b.x < 0 || b.y < 0 || b.x > m || b.y >n) return 1;
return 0;
}
void bfs()
{
priority_queue<Node>que;
Node a, b;
vis[0][0][0] = true;
a.x = a.y = a.enr = 0;
que.push(a);
while (!que.empty())
{
a = que.top();
que.pop();
for (int i = 0; i < 5; i++)
{
b = a;
b.x += dir[i][0];
b.y += dir[i][1];
b.enr++;
if (check(b)) continue;
if (vis[b.x][b.y][b.enr]) continue;
if (map[b.x][b.y]) continue;
if (b.enr + geth(b) > d) continue;
int dx,dy;
int flag = 1;
dx=b.x-1;
while (dx >= 0)//向北面找看看有没有往南面射箭的堡垒
{
if (map[dx][b.y])
{
int c = map[dx][b.y];//获取该堡垒的信息
int dis = b.x - dx;
if (cas[c].di == 'S')
{
if (dis%cas[c].v == 0)//计算这个点是否是箭的整数点的航行路线上
{
int time = dis / cas[c].v;
if (b.enr == time)
{
flag = 0;
break;
}
if (b.enr - time < 0) break;
if ((b.enr - time) % cas[c].t == 0)
{
flag = 0;
break;
}
else break;
}
else break;
}
else break;
}
dx--;
}
if (!flag) continue;
dx = b.x+1;
while (dx <= m)
{
if (map[dx][b.y])
{
int c = map[dx][b.y];
if (cas[c].di != 'N') break;
int dis = dx - b.x;
if (dis%cas[c].v == 0)
{
int time = dis / cas[c].v;
if(b.enr==time)
{
flag=0;
break;
}
if (b.enr- time < 0) break;
if ((b.enr - time) % cas[c].t == 0)
{
flag = 0;
break;
}
else break;
}
else break;
}
dx++;
}
if (!flag) continue;
dy=b.y+1;
while (dy <= n)
{
if (map[b.x][dy])
{
int c = map[b.x][dy];
if (cas[c].di != 'W') break;
int dis = dy - b.y;
if (dis%cas[c].v == 0)
{
int time = dis / cas[c].v;
if (b.enr == time)
{
flag = 0;
break;
}
if (b.enr - time < 0) break;
if ((b.enr - time) % cas[c].t == 0)
{
flag = 0;
break;
}
else break;
}
else break;
}
dy++;
}
if (!flag) continue;
dy = b.y-1;
while (dy >= 0)
{
if (map[b.x][dy])
{
int c = map[b.x][dy];
if (cas[c].di != 'E') break;
int dis = b.y - dy;
if (dis%cas[c].v == 0)
{
int time = dis / cas[c].v;
if (b.enr == time)
{
flag = 0;
break;
}
if (b.enr - time < 0)break;
if ((b.enr - time) % cas[c].t == 0)
{
flag = 0;
break;
}
else break;
}
else break;
}
dy--;
}
if (!flag) continue;
vis[b.x][b.y][b.enr] = true;
if (b.x == m && b.y == n)
{
printf("%d\n", b.enr);
return;
}
que.push(b);
}
}
printf("Bad luck!\n");
}
int main()
{
while (scanf("%d %d %d %d", &m, &n, &k, &d) != EOF)
{
memset(vis, false, sizeof vis);
memset(map, 0, sizeof(map));
for (int i = 1; i <= k; i++)
{
int a, b, c, d;
char e;
//getchar();
//scanf("%c %d %d %d %d", &e, &a, &b, &c, &d);
cin>>e>>a>>b>>c>>d;
map[c][d] = i;
cas[i].t = a, cas[i].v = b;
cas[i].di = e;
}
bfs();
}
//system("pause");
return 0;
}