洛谷P1608路径统计

题目

这个提示一个简单的最短路计数，除了用数组存上最短路的个数的做法以外，还有可以在得出最短路之后，搜索加剪枝的方法来通过该题。

可以反向搜索用A*的方法来通过，但是这个题的去重十分的恶心，需要一些玄学操作。

\(Code\)

// luogu-judger-enable-o2
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstdlib>
using namespace std;
struct cym {
	int to, len, nex;
}e[6000010], ed[6000010];

int lin[100100], lind[100100], vis[100100], dis[100100], dist[100100], map[2010][2010], n, m, minn = 2147483647, ans, cnt;
inline void add(int f, int t, int l)
{				
	e[++cnt].to = t;
	e[cnt].len = l;
	e[cnt].nex = lin[f];
	lin[f] = cnt;
	ed[cnt].to = f;
	ed[cnt].len = l;
	ed[cnt].nex = lind[t];
	lind[t] = cnt;
}	
void dfs(int now, int sum)
{
	if (now == n && sum == minn)
	{
		ans++;
		return;
	}
	if (now == n || dist[now] + sum > minn)
		return;
	if (sum >= minn) return;
	for (int i = lin[now]; i; i = e[i].nex)
	{
		dfs(e[i].to, sum + e[i].len);	
	}	
}
inline void spfa()
{	
	queue <int> q;
	for (int i = 2; i <= n; i++)
		dis[i] = 2147483647;
	dis[1] = 0;
	q.push(1);
	while (!q.empty())
	{
		int cur = q.front();
		q.pop(), vis[cur] = 0;
		for (int i = lin[cur]; i; i = e[i].nex)
		{
			int to = e[i].to;
			if (dis[cur] + e[i].len < dis[to])
			{
				dis[to] = dis[cur] + e[i].len;
				if (!vis[to])
					vis[to] = 1, q.push(to);
			}
		}
	}
}	
inline void spfad()
{
	memset(vis, 0, sizeof(vis));
	queue <int> q;
	for (int i = 1; i <= n; i++)
		dist[i] = 2147483647;
	dist[n] = 0;
	q.push(n);
	while (!q.empty())
	{
		int cur = q.front();
		q.pop(), vis[cur] = 0;
		for (int i = lind[cur]; i; i = ed[i].nex)
		{
			int to = ed[i].to;
			if (dist[cur] + ed[i].len < dist[to])
			{
				dist[to] = dist[cur] + ed[i].len;
				if (!vis[to])
					vis[to] = 1, q.push(to);
			}
		}
	}
}
int main()
{	
//	freopen("hh.txt", "r", stdin);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++)
	{
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if (map[a][b] == 0)
			map[a][b] = 214746;
		if (map[a][b] <= c) continue;
		add(a, b, c);
		map[a][b] = c;
	}
	spfa();
	spfad();
	/*for (int i = 1; i <= n; i++)
	{
		printf("%d ", dist[i]);
	}
	return 0;*/
	minn = dis[n];
	dfs(1, 0);
	if (m == 0 || minn == 2147483647)
		printf("No answer"), exit(0);
	printf("%d %d", minn, ans);
}