洛谷1462(重题1951) 通往奥格瑞玛的道路(收费站_NOI导刊2009提高(2))

1462原题链接

1951原题链接

显然答案有单调性，所以可以二分答案，用(SPFA)或(dijkstra)跑最短路来判断是否可行即可。
注意起点也要收费，(1462)数据较水，我一开始没判也过了，但重题(1951)把我卡掉了。。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 1e4 + 10;
const int M = 1e5 + 10;
struct dd {
	int x, D;
	bool operator < (const dd &b) const { return D > b.D; }
};
int fi[N], di[M], ne[M], da[M], dis[N], MI[N], f[N], l, n, HP, st, ed;
bool v[N];
priority_queue<dd>q;
inline 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, int z)
{
	di[++l] = y;
	da[l] = z;
	ne[l] = fi[x];
	fi[x] = l;
}
inline int maxn(int x, int y){ return x > y ? x : y; }
inline int minn(int x, int y){ return x < y ? x : y; }
bool dij(int V)
{
	if (f[st] > V)
		return false;
	int i, x, y;
	memset(dis, 60, sizeof(dis));
	memset(v, 0, sizeof(v));
	dis[st] = 0;
	q.push((dd){st, 0});
	while (!q.empty())
	{
		x = q.top().x;
		q.pop();
		if (v[x])
			continue;
		v[x] = 1;
		for (i = fi[x]; i; i = ne[i])
			if (dis[y = di[i]] > dis[x] + da[i] && f[di[i]] <= V)
				q.push((dd){y, dis[y] = dis[x] + da[i]});
	}
	return dis[ed] < HP;
}
int main()
{
	int i, m, x, y, z, l = 1e9, r = 0, mid;
	n = re();
	m = re();
	st = re();
	ed = re();
	HP = re();
	for (i = 1; i <= n; i++)
	{
		f[i] = re();
		l = minn(l, f[i]);
		r = maxn(r, f[i]);
	}
	for (i = 1; i <= m; i++)
	{
		x = re();
		y = re();
		z = re();
		add(x, y, z);
		add(y, x, z);
	}
	if (!dij(1e9))
	{
		printf("-1");
		return 0;
	}
	while (l < r)
	{
		mid = (l + r) >> 1;
		dij(mid) ? r = mid : l = mid + 1;
	}
	printf("%d", r);
	return 0;
}