洛谷1073 最优贸易

最短路

原题链接

以(1)为起点在正图上跑(SPFA)或(Dijkstra)，求出(dis1[x])，表示从(1)到节点(x)的所有路径中，经过权值最小的节点的权值；再以(n)为起点在反图上跑(SPFA)或(Dijkstra)，求出(dis2[x])，表示从(n)到节点(x)的所有路径中，经过权值最大的节点的权值。
最后枚举节点，求(max{dis2[x]-dis1[x]})即可。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1e5 + 10;
const int M = 5e5 + 10;
int fi[N], di[M << 1], pr[N], ne[M << 1], dis[N], disf[N], fif[N], dif[M << 1], nef[M << 1], q[M << 1], l, lf;
bool v[N];
int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c<'0' || c>'9'; c = getchar())
		p |= c == '-';
	for (; c >= '0'&&c <= '9'; c = getchar())
		x = x * 10 + (c - '0');
	return p ? -x : x;
}
inline void add(int x, int y)
{
	di[++l] = y;
	ne[l] = fi[x];
	fi[x] = l;
}
inline void addf(int x, int y)
{
	dif[++lf] = y;
	nef[lf] = fif[x];
	fif[x] = lf;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
inline int maxn(int x, int y)
{
	return x > y ? x : y;
}
int main()
{
	int i, n, m, x, y, head = 0, tail = 1, ma = 0;
	n = re();
	m = re();
	for (i = 1; i <= n; i++)
		pr[i] = re();
	for (i = 1; i <= m; i++)
	{
		x = re();
		y = re();
		if (re() == 1)
		{
			add(x, y);
			addf(y, x);
		}
		else
		{
			add(x, y);
			add(y, x);
			addf(x, y);
			addf(y, x);
		}
	}
	memset(dis, 60, sizeof(dis));
	q[1] = 1;
	dis[1] = pr[1];
	while (head != tail)
	{
		x = q[++head];
		v[x] = 0;
		for (i = fi[x]; i; i = ne[i])
		{
			y = di[i];
			if (dis[y] > minn(dis[x], pr[y]))
			{
				dis[y] = minn(dis[x], pr[y]);
				if (!v[y])
				{
					q[++tail] = y;
					v[y] = 1;
				}
			}
		}
	}
	memset(v, 0, sizeof(v));
	head = 0;
	tail = 1;
	q[1] = n;
	disf[n] = pr[n];
	while (head != tail)
	{
		x = q[++head];
		v[x] = 0;
		for (i = fif[x]; i; i = nef[i])
		{
			y = dif[i];
			if (disf[y] < maxn(disf[x], pr[y]))
			{
				disf[y] = maxn(disf[x], pr[y]);
				if (!v[y])
				{
					q[++tail] = y;
					v[y] = 1;
				}
			}
		}
	}
	for (i = 1; i <= n; i++)
		ma = maxn(ma, disf[i] - dis[i]);
	printf("%d", ma);
	return 0;
}