[洛谷P1951]收费站_NOI导刊2009提高（2）

题目大意：有一张$n$个点$m$条边的图，每个点有一个权值$w_i$，有边权，询问从$S$到$T$的路径中，边权和小于$s$，且$maxlimits_{路径经过k}{w_i}$最小，输出这个最小值，若到达不了，输出$-1$

题解：看到最大值最小，想到二分答案，二分这个最大值，每次对这个二分的答案跑一遍最短路，看是否可以到达就行了

卡点：1.没有判断起点的权值大于二分答案的情况

C++ Code：

#include <cstdio>
#include <algorithm>
#include <ext/pb_ds/priority_queue.hpp>
#define maxn 10010
#define maxm 50010
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int n, m, S, T, s, ans = -1;
int f[maxn], rnk[maxn];

inline bool cmp(int a, int b) {return f[a] < f[b];}

int head[maxn], cnt;
struct Edge {
	int to, nxt, w;
} e[maxm << 1];
void add(int a, int b, int c) {
	e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
}

long long d[maxn];
struct cmpq {
	inline bool operator () (const int &a, const int &b) const {
		return d[a] > d[b];
	}
};
__gnu_pbds::priority_queue<int, cmpq> q;
__gnu_pbds::priority_queue<int, cmpq>::point_iterator iter[maxn];
bool check(int mid) {
	if (f[S] > mid) return false;
	while (!q.empty()) q.pop();
	for (int i = 1; i <= n; i++) d[i] = inf, iter[i] = q.push(i);
	d[S] = 0;
	q.modify(iter[S], S);
	while (!q.empty()) {
		int u = q.top(); q.pop();
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if (f[v] > mid) continue;
			if (d[v] > d[u] + e[i].w) {
				d[v] = d[u] + e[i].w;
				if (d[T] <= s) return true;
				q.modify(iter[v], v);
			}
		}
	}
	return d[T] <= s;
}
int main() {
	scanf("%d%d%d%d%d", &n, &m, &S, &T, &s);
	for (int i = 1; i <= n; i++) scanf("%d", &f[i]), rnk[i] = i;
	for (int i = 1; i <= m; i++) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if (a == b) continue;
		add(a, b, c);
		add(b, a, c);
	}
	sort(rnk + 1, rnk + n + 1, cmp);
	int L = 1, R = n;
	while (L <= R) {
		int mid = L + R >> 1;
		if (check(f[rnk[mid]])) {
			ans = mid;
			R = mid - 1;
		} else L = mid + 1;
	}
	if (~ans) printf("%d
", f[rnk[ans]]);
	else puts("-1");
	return 0;
}