题意：
给定无向图，每条边有最大容量。从点1到点n的一条路径的容量取决于这条路径上容量最小的边的容量。现从点1到点n运输，求容量最大的路径的容量。
思路：
1. 令w[i]表示从点1到点i运输的最大流量，则w[i] = max(min(w[j], cost[j][i]), (1 <= j <= n))。可以用类似于最短路的算法来计算，如 (1)spfa (2)dijkstra (3)堆优化的dijkstra等。
2. 最大生成树。即求满足使1~n连通的容量最大的边的容量。
3. 二分 + dfs。
实现：
1.
(1)
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <queue> 5 using namespace std; 6 7 const int MAXN = 1005; 8 const int INF = 0x3f3f3f3f; 9 10 bool in[MAXN]; 11 int w[MAXN], T, n, m, x, y, c; 12 int G[MAXN][MAXN]; 13 14 int spfa() 15 { 16 queue<int> q; 17 q.push(1); 18 while (!q.empty()) 19 { 20 int tmp = q.front(); 21 q.pop(); 22 in[tmp] = false; 23 for (int i = 1; i <= n; i++) 24 { 25 if (min(w[tmp], G[tmp][i]) > w[i]) 26 { 27 w[i] = min(w[tmp], G[tmp][i]); 28 if (!in[i]) 29 { 30 in[i] = true; 31 q.push(i); 32 } 33 } 34 } 35 } 36 return w[n]; 37 } 38 int main() 39 { 40 cin >> T; 41 for (int t = 1; t <= T; t++) 42 { 43 cin >> n >> m; 44 for (int i = 1; i <= n; i++) 45 { 46 w[i] = (i == 1 ? INF : 0); 47 in[i] = (i == 1 ? true : false); 48 for (int j = 1; j <= n; j++) 49 { 50 G[i][j] = 0; 51 } 52 } 53 for (int i = 0; i < m; i++) 54 { 55 scanf("%d %d %d", &x, &y, &c); 56 G[x][y] = G[y][x] = c; 57 } 58 cout << "Scenario #" << t << ":" << endl; 59 cout << spfa() << endl << endl; 60 } 61 return 0; 62 }
(2)
#include <iostream> #include <cstdio> #include <vector> #include <queue> using namespace std; const int MAXN = 1005; const int INF = 0x3f3f3f3f; bool vis[MAXN]; int w[MAXN], T, n, m, x, y, c; int G[MAXN][MAXN]; int dijkstra() { for (int i = 1; i < n; i++) { int maxn = 0; int index = -1; for (int j = 1; j <= n; j++) { if (!vis[j] && w[j] > maxn) { maxn = w[j]; index = j; } } if (index == -1) break; vis[index] = true; for (int j = 1; j <= n; j++) { if (!vis[j] && min(w[index], G[index][j]) > w[j]) { w[j] = min(w[index], G[index][j]); } } } return w[n]; } int main() { cin >> T; for (int t = 1; t <= T; t++) { cin >> n >> m; for (int i = 1; i <= n; i++) { w[i] = (i == 1 ? INF : 0); vis[i] = false; for (int j = 1; j <= n; j++) { G[i][j] = 0; } } for (int i = 0; i < m; i++) { scanf("%d %d %d", &x, &y, &c); G[x][y] = G[y][x] = c; } cout << "Scenario #" << t << ":" << endl; cout << dijkstra() << endl << endl; } return 0; }
(3)
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <queue> 5 using namespace std; 6 typedef pair<int, int> P; 7 const int MAXN = 1005; 8 const int INF = 0x3f3f3f3f; 9 10 bool vis[MAXN]; 11 int w[MAXN], T, n, m, x, y, c; 12 int G[MAXN][MAXN]; 13 14 int dijkstra_heap() 15 { 16 priority_queue<P> q; 17 q.push(P(INF, 1)); 18 while (!q.empty()) 19 { 20 P tmp = q.top(); 21 q.pop(); 22 int now = tmp.second; 23 if (tmp.first < w[now]) 24 continue; 25 for (int i = 1; i <= n; i++) 26 { 27 if (min(w[now], G[now][i]) > w[i]) 28 { 29 w[i] = min(w[now], G[now][i]); 30 q.push(P(w[i], i)); 31 } 32 } 33 } 34 return w[n]; 35 } 36 int main() 37 { 38 cin >> T; 39 for (int t = 1; t <= T; t++) 40 { 41 cin >> n >> m; 42 for (int i = 1; i <= n; i++) 43 { 44 w[i] = (i == 1 ? INF : 0); 45 vis[i] = false; 46 for (int j = 1; j <= n; j++) 47 { 48 G[i][j] = 0; 49 } 50 } 51 for (int i = 0; i < m; i++) 52 { 53 scanf("%d %d %d", &x, &y, &c); 54 G[x][y] = G[y][x] = c; 55 } 56 cout << "Scenario #" << t << ":" << endl; 57 cout << dijkstra_heap() << endl << endl; 58 } 59 return 0; 60 }
2.
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <queue> 5 #include <algorithm> 6 using namespace std; 7 8 const int MAXN = 1005; 9 10 int T, n, m, x, y, c; 11 12 struct edge 13 { 14 int a, b, cost; 15 }; 16 edge es[1000005]; 17 int ran[MAXN]; 18 int par[MAXN]; 19 20 void init(int n) 21 { 22 for (int i = 0; i < n; i++) 23 { 24 par[i] = i; 25 ran[i] = 0; 26 } 27 } 28 29 int find(int x) 30 { 31 if (par[x] == x) 32 return x; 33 return par[x] = find(par[x]); 34 } 35 36 void unite(int x, int y) 37 { 38 x = find(x); 39 y = find(y); 40 if (x == y) 41 return; 42 if (ran[x] < ran[y]) 43 { 44 par[x] = y; 45 } 46 else 47 { 48 par[y] = x; 49 if (ran[x] == ran[y]) 50 { 51 ran[x] ++; 52 } 53 } 54 } 55 56 bool same(int x, int y) 57 { 58 return find(x) == find(y); 59 } 60 61 bool cmp(edge a, edge b) 62 { 63 return a.cost > b.cost; 64 } 65 66 int kruskal() 67 { 68 init(n); 69 sort(es, es + m, cmp); 70 for (int i = 0; i < m; i++) 71 { 72 if (!same(es[i].a, es[i].b)) 73 { 74 unite(es[i].a, es[i].b); 75 } 76 if (same(0, n - 1)) 77 return es[i].cost; 78 } 79 return es[m - 1].cost; 80 } 81 82 int main() 83 { 84 cin >> T; 85 for (int t = 1; t <= T; t++) 86 { 87 cin >> n >> m; 88 for (int i = 0; i < m; i++) 89 { 90 scanf("%d %d %d", &x, &y, &c); 91 es[i].a = x - 1; 92 es[i].b = y - 1; 93 es[i].cost = c; 94 } 95 cout << "Scenario #" << t << ":" << endl; 96 cout << kruskal() << endl << endl; 97 } 98 return 0; 99 }
3.
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 const int MAXN = 1005; 8 9 bool vis[MAXN]; 10 int T, n, m, x, y, c; 11 int G[MAXN][MAXN]; 12 13 bool dfs(int now, int x) 14 { 15 if (now == n) 16 return true; 17 vis[now] = true; 18 for (int i = 1; i <= n; i++) 19 { 20 if (G[now][i] >= x && !vis[i]) 21 if (dfs(i, x)) 22 return true; 23 } 24 return false; 25 } 26 27 bool check(int x) 28 { 29 memset(vis, 0, sizeof(vis)); 30 return dfs(1, x); 31 } 32 33 int search(int maxn) 34 { 35 int l = 1, r = maxn, m, res = 0; 36 while (l <= r) 37 { 38 int m = (l + r) >> 1; 39 if (check(m)) 40 { 41 l = m + 1; 42 res = m; 43 } 44 else 45 { 46 r = m - 1; 47 } 48 } 49 return res; 50 } 51 int main() 52 { 53 cin >> T; 54 for (int t = 1; t <= T; t++) 55 { 56 scanf("%d %d", &n, &m); 57 memset(G, 0, sizeof(G)); 58 int maxn = 0; 59 for (int i = 0; i < m; i++) 60 { 61 scanf("%d %d %d", &x, &y, &c); 62 G[x][y] = G[y][x] = c; 63 maxn = max(maxn, c); 64 } 65 cout << "Scenario #" << t << ":" << endl; 66 cout << search(maxn) << endl << endl; 67 } 68 return 0; 69 }