http://poj.org/problem?id=2455
题意：给出n个点和m条无向路，每条路都有一个长度。从1点到n点要走t次两两互不重合的路。求出每条1->n的路中相邻两点最大值的最小值。
思路：题目就是要最小化最大值，因此可以二分枚举当前的最大长度，如果长度小于等于当前枚举的值的话，就可以给它容量，否则就设容量为0，然后跑一遍最大流判断是否流量大于等于t，根据情况更新图。
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 #define N 410 6 #define INF 0x3f3f3f3f 7 struct Edge { 8 int v, nxt, cap, w; 9 Edge () {} 10 Edge (int v, int nxt, int cap, int w) : v(v), nxt(nxt), cap(cap), w(w) {} 11 } edge[N*N]; 12 int head[N], tot, dis[N], cur[N], pre[N], gap[N], n, m, t; 13 14 void Add(int u, int v, int cap, int w) { 15 edge[tot] = Edge(v, head[u], cap, w); head[u] = tot++; 16 edge[tot] = Edge(u, head[v], 0, w); head[v] = tot++; 17 } 18 19 int BFS(int S, int T) { 20 queue<int> que; que.push(T); 21 memset(dis, INF, sizeof(dis)); 22 memset(gap, 0, sizeof(gap)); 23 gap[0]++; dis[T] = 0; 24 while(!que.empty()) { 25 int u = que.front(); que.pop(); 26 for(int i = head[u]; ~i; i = edge[i].nxt) { 27 int v = edge[i].v; 28 if(dis[v] == INF) { 29 dis[v] = dis[u] + 1; 30 gap[dis[v]]++; 31 que.push(v); 32 } 33 } 34 } 35 } 36 37 int ISAP(int S, int T, int n) { 38 BFS(S, T); 39 memcpy(cur, head, sizeof(cur)); 40 int u = pre[S] = S, i, index, flow, ans = 0; 41 while(dis[S] < n) { 42 if(u == T) { 43 flow = INF, index = S; // index = S !!! 44 for(i = S; i != T; i = edge[cur[i]].v) 45 if(flow > edge[cur[i]].cap) flow = edge[cur[i]].cap, index = i; 46 for(i = S; i != T; i = edge[cur[i]].v) 47 edge[cur[i]].cap -= flow, edge[cur[i]^1].cap += flow; 48 ans += flow, u = index; 49 } 50 for(i = cur[u]; ~i; i = edge[i].nxt) 51 if(edge[i].cap > 0 && dis[edge[i].v] == dis[u] - 1) break; 52 if(~i) { 53 pre[edge[i].v] = u; cur[u] = i; u = edge[i].v; 54 } else { 55 if(--gap[dis[u]] == 0) break; 56 int md = n; 57 for(i = head[u]; ~i; i = edge[i].nxt) 58 if(md > dis[edge[i].v] && edge[i].cap > 0) md = dis[edge[i].v], cur[u] = i; 59 gap[dis[u] = md + 1]++; 60 u = pre[u]; 61 } 62 } 63 return ans; 64 } 65 66 int main() { 67 while(~scanf("%d%d%d", &n, &m, &t)) { 68 int r = 0, l = INF; 69 memset(head, -1, sizeof(head)); tot = 0; 70 for(int i = 1; i <= m; i++) { 71 int u, v, w; scanf("%d%d%d", &u, &v, &w); 72 Add(u, v, 0, w); 73 if(r < w) r = w; 74 if(l > w) l = w; 75 } 76 int ans = r; 77 while(l <= r) { 78 int mid = (l + r) >> 1; 79 for(int i = 0; i < tot; i++) 80 if(edge[i].w <= mid) edge[i].cap = 1; 81 else edge[i].cap = 0; 82 if(ISAP(1, n, n) >= t) r = mid - 1, ans = mid; 83 else l = mid + 1; 84 } 85 printf("%d ", ans); 86 } 87 return 0; 88 }