題目大意：有一張$n$個(gè)點(diǎn)$m$條邊的圖，每個(gè)點(diǎn)有一個(gè)權(quán)值$w_i$，有邊權(quán)，詢問從$S$到$T$的路徑中，邊權(quán)和小于$s$，且$\max\limits_{路徑經(jīng)過(guò)k}\{w_i\}$最小，輸出這個(gè)最小值，若到達(dá)不了，輸出$-1$

題解：看到最大值最小，想到二分答案，二分這個(gè)最大值，每次對(duì)這個(gè)二分的答案跑一遍最短路，看是否可以到達(dá)就行了

卡點(diǎn)：1.沒有判斷起點(diǎn)的權(quán)值大于二分答案的情況

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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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Memory-of-winter/p/9498773.html

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