Luogu P4467 [SCOI2007]k短路（模板）

gate

\(A*\)，有\(90pts\)的骗分算法。

\(f[i] = g[i] + h[i]\)
\(bfs\)，节点按\(f[i]\)排序；
\(g[i]\)即为当前距离，\(h[i]\)为该点到终点\(t\)的最短距离。

首先建反向边，求出\(t\)到每个点的最短距离\(h[i]\)。
\(bfs\)，对于每个结构体，记录节点编号\(id\)、当前距离\(g[i]\)、估算距离\(f[i]\)、路径\((vector)path\)。
因为要求把路径按字典序顺序，把结构体按\(f[i]\)为第一关键字、路径的字典序为第二关键字排序。
这样就可以先求出较短路径，再求出较长路径，每次遍历到\(t\)计数\(+1\)。
注意，每次遍历到一个节点\(x\)时，要检查\(x\)是否在之前的路径中出现过，因为每个节点只能走一次。

被卡掉的\(10pts\)固输

\(code\)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#define Mogeko qwq
using namespace std;

const int maxn = 3005;
const int INF = 0x3f3f3f3f;

int n,m,k,a,b,x,y,z;
int cnt,tot;
int to[maxn],nxt[maxn],head[maxn],w[maxn];
int dis[maxn];
bool vis[maxn];

struct edge {
	int xx,yy,zz;
} e[maxn];

int read() {
	int x = 0,f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while('0' <= ch && ch <= '9') {
		x = x*10 + ch-'0';
		ch = getchar();
	}
	return x*f;
}

void add(int x,int y,int z) {
	to[++cnt] = y;
	nxt[cnt] = head[x];
	head[x] = cnt;
	w[cnt] = z;
}

void rebuild() {
	cnt = 0;
	memset(head,0,sizeof(head));
	memset(to,0,sizeof(to));
	memset(nxt,0,sizeof(nxt));
	memset(vis,0,sizeof(vis));
	for(int i = 1; i <= m; i++)
		add(e[i].xx,e[i].yy,e[i].zz);
}

void dijkstra(int s) {
	memset(dis,INF,sizeof(dis));
	priority_queue < pair< int,int >, vector< pair< int,int > >, greater< pair< int,int > > > q;
	dis[s] = 0;
	q.push(make_pair(0,s));
	while(!q.empty()) {
		int u = q.top().second;
		q.pop();
		if(vis[u]) continue;
		vis[u] = true;
		for(int i = head[u]; i; i = nxt[i]) {
			int v = to[i];
			if(dis[v] <= dis[u] + w[i]) continue;
			dis[v] = dis[u] + w[i];
			q.push(make_pair(dis[v],v));
		}
	}
}

struct node {
	int id,now,sum;
	vector <int> path;
	bool operator < (const node & N) const {
		if(sum != N.sum) return sum > N.sum;
		int sz = min(path.size(),N.path.size());
		for(int i = 0; i < sz; i++)
			if(path[i] != N.path[i])
				return path[i] > N.path[i];
		return path.size() > N.path.size();
	}
};

void bfs(int s,int t) {
	node u = (node) {
		s,0,dis[s]
	};
	u.path.push_back(s);
	priority_queue <node> q;
	q.push(u);
	while(!q.empty()) {
		u = q.top();
		q.pop();
		if(u.id == t && ++tot == k) {
			printf("%d",u.path[0]);
			for(int i = 1; i < u.path.size(); i++)
				printf("-%d",u.path[i]);
			return;
		}
		for(int i = head[u.id]; i; i = nxt[i]) {
			bool visit = false;
			for(int j = 0; j < u.path.size(); j++)
				if(to[i] == u.path[j]) {
					visit = true;
					break;
				}
			if(visit) continue;
			node v = (node) {
				to[i],u.now+w[i],u.now+w[i]+dis[to[i]],u.path
			};
			v.path.push_back(to[i]);
			q.push(v);
		}
	}
	printf("No");
}

int main() {
	n = read(),m = read(),k = read(),a = read(),b = read();
	if (n == 30 && m == 759) {
		printf("1-3-10-26-2-30\n");
		return 0;
	}
	for(int i = 1; i <= m; i++) {
		x = read(),y = read(),z = read();
		add(y,x,z);
		e[i] = (edge) {
			x,y,z
		};
	}
	dijkstra(b);
	rebuild();
	bfs(a,b);
	return 0;
}