「BZOJ 1698」「USACO 2007 Feb」Lilypad Pond 荷叶池塘「最短路」

题解

从一个点P可以跳到另一个点Q，如果Q是水这条边就是1，如果Q是荷叶这条边权值是0。可以跑最短路并计数

问题是边权为0的最短路计数没有意义（只是荷叶的跳法不同），所以我们两个能通过荷叶间接连通的点连一条边权为1的边就好

#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;

#define id(x, y) ((x - 1) * m + y)
#define valid(x, y) (x >= 1 && y >= 1 && x <= n && y <= m)
#define pt(x, y) (a[x][y] == 1 || a[x][y] == 3 || a[x][y] == 4)

const int N = 35;
const int dx[] = {1, 1, -1, -1, 2, 2, -2, -2};
const int dy[] = {2, -2, 2, -2, 1, -1, 1, -1};

struct Edge {
	int v, nxt;
} e[N * N * N * N];
int n, m, a[N][N], hd[N * N], p;
bool edge[N * N][N * N];

void clr() {
	fill(hd + 1, hd + n * m + 1, -1); p = 0;
}
void add(int u, int v) {
	if(edge[u][v] || u == v) return ;
	edge[u][v] = 1;
	e[p] = (Edge) {v, hd[u]}; hd[u] = p ++;
}
bool vis[N * N];
void dfs(int x, int y, int cur) {
	vis[id(x, y)] = 1;
	for(int k = 0; k < 8; k ++) {
		int nx = x + dx[k], ny = y + dy[k];
		if(valid(nx, ny) && !vis[id(nx, ny)]) {
			if(a[nx][ny] == 1) dfs(nx, ny, cur);
			else add(cur, id(nx, ny));
		}
	}
}
int dis[N * N];
long long cnt[N * N]; 
void bfs(int s, int t) {
	fill(dis + 1, dis + n * m + 1, -1);
	queue<int> q; q.push(s); cnt[s] = 1; dis[s] = 0;
	while(q.size()) {
		int u = q.front(); q.pop();
		for(int i = hd[u]; ~ i; i = e[i].nxt) {
			int v = e[i].v;
			if(dis[v] == -1) {
				dis[v] = dis[u] + 1;
				cnt[v] = cnt[u]; q.push(v);
			} else if(dis[v] == dis[u] + 1) {
				cnt[v] += cnt[u];
			}
		}
	}
}
int main() {
	scanf("%d%d", &n, &m); clr();
	int s = 1, t = 1;
	for(int i = 1; i <= n; i ++) {
		for(int j = 1; j <= m; j ++) {
			scanf("%d", &a[i][j]);
			if(a[i][j] == 3) s = id(i, j);
			if(a[i][j] == 4) t = id(i, j);
		}
	}
	for(int i = 1; i <= n; i ++) {
		for(int j = 1; j <= m; j ++) if(a[i][j] == 3 || a[i][j] == 0) {
			fill(vis + 1, vis + n * m + 1, 0);
			dfs(i, j, id(i, j));
		}
	}
	bfs(s, t);
	if(cnt[t]) printf("%d
%lld
", dis[t] == 0 ? 0 : dis[t] - 1, cnt[t]);
	else puts("-1");
	return 0;
}