[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$\max\limits_{路径经过k}\{w_i\}$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\n", f[rnk[ans]]);else puts("-1");return 0;
}