RT，第 16 个点 WA 了。

思路是差分判断一个点是否能被看到，然后 bfs

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

const int MAXN = 360;
const int INF = 0x3f3f3f3f;

int n, m, hide, quick, d;
int a[MAXN][MAXN], b[MAXN][MAXN], c[MAXN][MAXN], sx, sy, tx, ty;
string in;

inline void watch(int x, int y, int dis)
{
	--dis;
	for(int i=0;i<=dis;i++)
	{
		int minx = max(1, x-i), maxx = min(n, x+i);
		int miny = max(1, y-dis+i), maxy = min(n, y+dis-i);
		++b[minx][miny]; --b[minx][maxy+1];
		++b[maxx][miny]; --b[maxx][maxy+1];
	}
}

int dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
int dy[] = {1, 0, -1, 0, -1, -1, 1, 1};

struct node{
	int nx, ny, step, nowh, nowq;
	node(int X=0,int Y=0,int S=INF,int H=INF,int Q=INF):nx(X),ny(Y),step(S),nowh(H),nowq(Q)
	{};
	bool operator<(const node &o)
	{
		if(step != o.step) return step<o.step;
		else if(nowh+nowq != o.nowh+o.nowq) return nowh+nowq<o.nowh+o.nowq;
		else return nowh<o.nowh;
	}
};
node q[20000005], ans;
int h = 1, t = 0;
bool vis[MAXN][MAXN][20][20];

inline bool inside(int x, int y)
{
	return x >= 1 && x <= n && y >= 1 && y <= m; 
}

inline void bfs()
{
	vis[sx][sy][0][0] = 1;
	q[++t] = node(sx, sy, 0, 0, 0);
	while(h <= t)
	{
		node now = q[h++];
		if(now.step > ans.step) continue;
		if(now.nx == tx && now.ny == ty)
		{
			if(now < ans) ans = now;
		}
		for(int i=0;i<8;i++)
		{
			int xx = now.nx+dx[i], yy = now.ny+dy[i];
			if(inside(xx, yy) && a[xx][yy] <= 0)
			{
				if(c[xx][yy])
				{
					if(now.nowh+1 <= hide && !vis[xx][yy][now.nowh+1][now.nowq])
					{
						vis[xx][yy][now.nowh+1][now.nowq] = 1;
						q[++t] = node(xx, yy, now.step+1, now.nowh+1, now.nowq); 
					}
				}
				else
				{
					if(!vis[xx][yy][now.nowh][now.nowq])
					{
						vis[xx][yy][now.nowh][now.nowq] = 1;
						q[++t] = node(xx, yy, now.step+1, now.nowh, now.nowq); 
					}
				}
			}
		}
		if(now.nowq+1 <= quick)
		{
			for(int i=0;i<4;i++)
			{
				int xx = now.nx + dx[i]*d, yy = now.ny + dy[i]*d;
				if(inside(xx, yy) && a[xx][yy] <= 0)
				{
					if(c[xx][yy])
					{
						if(now.nowh+1 <= hide && !vis[xx][yy][now.nowh+1][now.nowq+1])
						{
							vis[xx][yy][now.nowh+1][now.nowq+1] = 1;
							q[++t] = node(xx, yy, now.step+1, now.nowh+1, now.nowq+1); 
						}
					}
					else
					{
						if(!vis[xx][yy][now.nowh][now.nowq+1])
						{
							vis[xx][yy][now.nowh][now.nowq+1] = 1;
							q[++t] = node(xx, yy, now.step+1, now.nowh, now.nowq+1); 
						}
					}
				}
			}
		}
	}
}

int main()
{
	scanf("%d%d%d%d%d",&n,&m,&hide,&quick,&d);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cin>>in;
			if(in[0] == 'S')
			{
				a[i][j] = -1;
				sx = i; sy = j;
			}
			else if(in[0] == 'T')
			{
				a[i][j] = -2;
				tx = i; ty = j;
			}
			else if(in[0] == '.') a[i][j] = 0;
			else 
			{
				int now = 0;
				for(int k=0;k<in.size();k++)
					now = (now<<3)+(now<<1)+in[k]-'0';
				a[i][j] = now;
				watch(i, j, now);
			}
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			b[i][j] += b[i][j-1];
			if(b[i][j] > 0) c[i][j] = 1;
		}
	bfs();
	if(ans.step == INF) puts("-1");
	else printf("%d %d %d\n",ans.step, ans.nowh, ans.nowq);
	return 0;
}

希望大佬们看得惯我的码风 /kel

但是大括号换行的人似乎不多

另外我英语不好有些变量名可能很奇怪QAQ