题目来源：晴天的魔法乐园
分析：
综合上述讨论，本题的做法就是，先求一次最短路，用pre数组记录【其中一条】最短路径，然后遍历这条最短路径上的边，删除后各求一次路径，取他们中从起点到终点的最短距离的最大值即可
1 //// patest.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。 2 //// 3 #include "pch.h" 4 5 #include <cstdio> 6 #include <cmath> 7 #include <cstdlib> 8 #include <cstring> 9 #include <iostream> 10 #include <iterator> 11 #include <algorithm> 12 #include <string> 13 #include <set> 14 #include <unordered_set> 15 #include <map> 16 #include <unordered_map> 17 #include <vector> 18 #include <stack> 19 #include <queue> 20 #include <deque> 21 22 using namespace std; 23 const int INF = 0x7fffffff; 24 const int maxn = 1010; 25 const int maxk = 10010; 26 27 //void printVector(vector<int>& vt) { 28 // int size = vt.size(); 29 // for (int i = 0; i < size; i++) { 30 // printf("%d", vt[i]); 31 // if (i != size - 1) printf(" "); 32 // } 33 // printf("\n"); 34 // 35 //} 36 37 38 int n, m, st, ed; 39 bool vis[maxn]; 40 int d[maxn]; 41 int G[maxn][maxn]; 42 43 vector<int> pre[maxn]; 44 45 void Dijkstra(int s) { 46 47 memset(vis, false, sizeof(vis)); 48 fill(d, d + maxn, INF); 49 d[s] = 0; 50 for (int i = 0; i < n; i++) { 51 int u = -1, MIN = INF; 52 for (int j = 1; j <= n; j++) { 53 if (vis[j] == false && d[j] < MIN) { 54 MIN = d[j]; 55 u = j; 56 } 57 } 58 59 if (u == -1) return; 60 vis[u] = true; 61 for (int v = 1; v <= n; v++) { 62 if (vis[v] == false && G[u][v] != INF) { 63 if (d[u] + G[u][v] < d[v]) { 64 d[v] = d[u] + G[u][v]; 65 pre[v].clear(); 66 pre[v].push_back(u); 67 } 68 else if (d[u] + G[u][v] == d[v]) { 69 pre[v].push_back(u); 70 } 71 } 72 } 73 74 } 75 } 76 77 vector<pair<int, int>> ans; 78 79 void DFS(int index) { 80 for (auto next : pre[index]) { 81 ans.push_back(make_pair(index, next)); 82 DFS(next); 83 } 84 } 85 86 int maxTime = -1; 87 int main() { 88 89 fill(G[0], G[0] + maxn * maxn, INF); 90 91 scanf("%d%d", &n, &m); 92 st = 1, ed = n; 93 for (int i = 0; i < m; i++) { 94 int u, v, time; 95 scanf("%d%d%d", &u, &v, &time); 96 G[u][v] = G[v][u] = time; 97 98 } 99 100 Dijkstra(st); 101 DFS(ed); 102 103 int lastTime; 104 for (int i = 0; i < ans.size(); i++) { 105 int u = ans[i].first; 106 int v = ans[i].second; 107 if (i > 0) { 108 int id1 = ans[i - 1].first; 109 int id2 = ans[i - 1].second; 110 G[id1][id2] = G[id2][id1] = lastTime; 111 } 112 113 lastTime = G[u][v]; 114 G[u][v] = G[v][u] = INF; 115 116 Dijkstra(st); 117 maxTime = max(maxTime, d[ed]); 118 119 } 120 121 if (maxTime == INF) printf("It's a bug!!!\n"); 122 else printf("%d\n", maxTime); 123 124 125 126 127 return 0; 128 }
