SPFA(负环) LightOJ 1074 Extended Traffic

题目传送门

题意:收过路费.如果最后的收费小于3或不能达到,输出'?'.否则输出到n点最小的过路费

分析:关键权值可为负,如果碰到负环是,小于3的约束条件不够,那么在得知有负环时,把这个环的点都标记下,DFS实现.

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

const int N = 2e2 + 5;
const int E = 1e4 + 5;
const int INF = 0x3f3f3f3f;
struct Edge	{
	int v, w, nex;
	Edge()	{}
	Edge(int v, int w, int nex) : v (v), w (w), nex (nex) {}
	bool operator < (const Edge &r) const	{
		return w > r.w;
	}
}edge[E];
int head[N];
int d[N];
int cnt[N];
int a[N];
bool vis[N], vis2[N];
int n, m, e;

void init()	{
	memset (head, -1, sizeof (head));
	e = 0;
}

void add_edge(int u, int v, int w)	{
	edge[e] = Edge (v, w, head[u]);
	head[u] = e++;
}

void DFS(int u)	{
	vis2[u] = true;
	for (int i=head[u]; ~i; i=edge[i].nex)	{
		int v = edge[i].v;
		if (!vis2[v])	{
			DFS (v);
		}
	}
}

void SPFA(int s)	{
	memset (cnt, 0, sizeof (cnt));
	memset (vis, false, sizeof (vis));
	memset (vis2, false, sizeof (vis2));
	memset (d, INF, sizeof (d));
	d[s] = 0;	cnt[s] = 0;	vis[s] = true;
	queue<int> que;	que.push (s);
	while (!que.empty ())	{
		int u = que.front ();	que.pop ();
		vis[u] = false;
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			int v = edge[i].v, w = edge[i].w;
			if (vis2[v])	continue;
			if (d[v] > d[u] + w)	{
				d[v] = d[u] + w;
				if (!vis[v])	{
					vis[v] = true;	que.push (v);
					if (++cnt[v] > n)	{
						DFS (v);
					}
				}
			}
		}
	}
}

int cal(int i, int j)	{
	int ret = a[i] - a[j];
	ret = ret * ret * ret;
	return ret;
}

int main(void)	{
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		init ();
		scanf ("%d", &n);
		for (int i=1; i<=n; ++i)	{
			scanf ("%d", &a[i]);
		}
		scanf ("%d", &m);
		for (int u, v, i=1; i<=m; ++i)	{
			scanf ("%d%d", &u, &v);
			add_edge (u, v, cal (v, u));
		}
		SPFA (1);
		int q;	scanf ("%d", &q);
		printf ("Case %d:
", ++cas);
		while (q--)	{
			int x;	scanf ("%d", &x);
			if (d[x] == INF || d[x] < 3 || vis2[x])	puts ("?");
			else	printf ("%d
", d[x]);
		}
	}

	return 0;
}