【luogu P1606 [USACO07FEB]荷叶塘Lilypad Pond】 题解

题目链接：https://www.luogu.org/problemnew/show/P1606

这个题。。第一问很好想，但是第二问，如果要跑最短路计数的话，零边权的花怎么办？

不如这样想，如果这个点能到花的话，那把他和从花能到的一个点边权连成一，好比两条路径共为1：一条为1一条为0的路径

但在实际操作的时候，一朵花是可以到另一朵花的！

电风扇好啊

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;
const int maxn = 2000000;
ll a[50][50], n, m, s, t, ans[maxn], dis[maxn];
bool vis[maxn], used[maxn];
ll dx[9] = {0,-2,-2,-1,+1,+2,+2,+1,-1}, dy[9] = {0,-1,+1,+2,+2,+1,-1,-2,-2};
queue<ll> q;
struct edge{
	ll to, next, len;
}e[maxn<<2];
ll head[maxn], cnt;
void add(ll u, ll v, ll w)
{
	e[++cnt].len = w; e[cnt].to = v; e[cnt].next = head[u]; head[u] = cnt; 
}
void SPFA()
{
	for(ll i = 1; i <= n*m; i++) dis[i] = 9223372036854775806;
	q.push(s);
	vis[s] = 1;
	dis[s] = 0;
	ans[s] = 1;
	while(!q.empty())
	{
		ll now = q.front(); q.pop();
		vis[now] = 0;
		for(ll i = head[now]; i != -1; i = e[i].next)
		{
			ll v = e[i].to;
			if(dis[v] > dis[now] + e[i].len)
			{
				ans[v] = ans[now];
				dis[v] = dis[now] + e[i].len;
				if(!vis[v])
				{
					q.push(v);
					vis[v] = 1;
				}
			}
			else if(dis[v] == dis[now] + e[i].len) ans[v] += ans[now];
		}
	}
}
void dfs(ll fx, ll fy, ll tx, ll ty)
{
	used[(tx-1)*m+ty] = 1;
	for(ll i = 1; i <= 8; i++)
	{
		ll ex = tx+dx[i], ey = ty+dy[i];
		if(ex < 1 || ex > n || ey < 1 || ey > m || used[(ex-1)*m+ey]) continue;
		if(a[ex][ey] == 2) continue;
		if(a[ex][ey] == 1) dfs(fx, fy, ex, ey);
		else used[(ex-1)*m+ey] = 1, add((fx-1)*m+fy, (ex-1)*m+ey, 1);
	}
}
void init()
{
	memset(head, -1, sizeof(head));
	scanf("%lld%lld",&n,&m);
	for(ll i = 1; i <= n; i++)
		for(ll j = 1; j <= m; j++) 
		{
			scanf("%lld",&a[i][j]);
			if(a[i][j] == 3)
			s = (i-1)*m+j;
			if(a[i][j] == 4)
			t = (i-1)*m+j;
		}
	for(ll i = 1; i <= n; i++)
		for(ll j = 1; j <= m; j++)
		{
			if(a[i][j] == 2 || a[i][j] == 4 || a[i][j] == 1) continue;
			memset(used, 0, sizeof(used));
			dfs(i, j, i, j);
		}
}
#undef ll
int main()
#define ll long long
{
	freopen("testdata.in","r",stdin);
	init();
	SPFA();
	if(dis[t] == 9223372036854775806) puts("-1");
	else printf("%lld
%lld",dis[t]-1, ans[t]);
	return 0;
}