题解【洛谷P1841】[JSOI2007]重要的城市

题面

题解

最短路图模板题。

介绍一下最短路图：

• 先对原图跑一边单源最短路,求出源点到每个点\(i\)的最短路\(dis[i]\).
• 接下来构建新图：对于一条边\((x,y,v)\),若\(dis[x]+v=dis[y]\)则在新图中加入这条边.
• 这样就构成了一个新图,满足\(DAG\)的性质

对于此题：

先建立最短路图，
考虑最短路图上的一条边\((x,y)\),若\(y\)的入度为\(1\),则\(x\)为重要的.

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <queue>

using namespace std;

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

int tot, n, cnt, dis[1000003], m, qi[1000003], head[1000003], in[3000003], nxt[3000003], ver[1000003], edge[1000003], ans[1000003], sum;
priority_queue <pair <int, int> > q;
int vis[1000003], inans[1000003], bj[100003];

inline void add(int u, int v, int w)
{
	ver[++cnt] = v, qi[cnt] = u, edge[cnt] = w, nxt[cnt] = head[u], head[u] = cnt;
}

inline void solve(int uuu)
{
	memset(dis, 0x3f3f3f3f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	q.push(make_pair(0, uuu));
	dis[uuu] = 0;
	while (!q.empty())
	{
		int u = q.top().second; q.pop();
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i], w = edge[i];
			if (dis[v] > dis[u] + w)
			{
				dis[v] = dis[u] + w;
				q.push(make_pair(-dis[v], v));
			}
		}
	}
	memset(bj, 0, sizeof(bj));
	memset(in, 0, sizeof(in));
	for (int i = 1; i <= cnt; i+=1)
	{
		int u = qi[i], v = ver[i], w = edge[i];
		if (dis[u] + w == dis[v])
		{
			bj[i] = 1;
			++in[v];
		}
	}
	for (int i = 1; i <= cnt; i+=1)
	{
		int u = qi[i], v = ver[i];
		if (bj[i] && in[v] == 1 && u != uuu)
		{
			inans[u] = 1;
		}
	}
}

int main()
{
	n = gi(), m = gi();
	for (int i = 1; i <= m; i++)
	{
		int u = gi(), v = gi(), w = gi();
		add(u, v, w), add(v, u, w);
	}
	for (int i = 1; i <= n; i+=1) solve(i);
	bool fl = true;
	for (int i = 1; i <= n; i+=1)
	{
		if (inans[i]) {fl = false; printf("%d ", i);}
	}
	if (fl) puts("No important cities.");
	return 0;
}