Luogu 1606 [USACO07FEB]白银莲花池Lilypad Pond

感觉应当挺简单的，但是弄了好久……菜死了

如果不考虑那些为$1$的点，直接跑个最短路计数就好了，但是我们现在有一些边可以不用付出代价，那么只要在连边的时候先预处理搜一下就好了。

原来的想法是拆点，但是这样子不好连边，所以直接把点权转化到边权上来。

注意到起点其实不用付出代价，那么最后的答案就是$dis_ed - 1$。

时间复杂度上界是$O(n^4 + n^2log(n^2))$。

Code：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair <ll, int> pin;

const int N = 1005;
const int M = 2e5 + 5;
const int L = 35;
const int dx[] = {-1, -2, -2, -1, 1, 2, 2, 1};
const int dy[] = {-2, -1, 1, 2, 2, 1, -1, -2};
const ll inf = 0x3f3f3f3f3f3f3f3f;

int n, m, a[L][L], tot = 0, head[N];
ll dis[N], cnt[N];
bool vis[N], ins[L][L];

struct Edge {
    int to, nxt;
} e[M];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline int id(int x, int y) {
    return (x - 1) * m + y;
}

void dfs(int nowId, int x, int y) {
    ins[x][y] = 1;
    for(int i = 0; i < 8; i++) {
        int tox = x + dx[i], toy = y + dy[i];
        if(tox < 1 || tox > n || toy < 1 || toy > m) continue;
        if(ins[tox][toy]) continue;
        if(a[tox][toy] == 1) dfs(nowId, tox, toy);
        else ins[tox][toy] = 1, add(nowId, id(tox, toy));
    }
}

priority_queue <pin> Q;
void dij(int st) {
    memset(dis, 0x3f, sizeof(dis));
    cnt[st] = 1LL, dis[st] = 0LL;
    Q.push(pin(0, st));
    for(; !Q.empty(); ) {
        int x = Q.top().second; Q.pop();
        if(vis[x]) continue;
        vis[x] = 1;
        for(int i = head[x]; i; i = e[i].nxt) {
            int y = e[i].to;
            if(dis[y] > dis[x] + 1) {
                dis[y] = dis[x] + 1;
                cnt[y] = cnt[x];
                Q.push(pin(-dis[y], y));
            } else if(dis[y] == dis[x] + 1) {
                cnt[y] += cnt[x];
            }
        }
    } 
}

int main() {
//    freopen("testdata.in", "r", stdin);
    
    read(n), read(m);
    int st, ed;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            read(a[i][j]);
            if(a[i][j] == 3) st = id(i, j);
            if(a[i][j] == 4) ed = id(i, j);
        }
    
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(a[i][j] == 0 || a[i][j] == 3) {
                memset(ins, 0, sizeof(ins));
                dfs(id(i, j), i, j);
            } 
    
    dij(st);
    
    if(dis[ed] == inf) puts("-1");
    else printf("%lld
%lld
", dis[ed] - 1, cnt[ed]);
    
    return 0;
}