POJ 2455：Secret Milking Machine（二分+最大流）

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。

 #include <cstdio>
 #include <cstring>
 #include <queue>
 using namespace std;
 #define N 410
 #define INF 0x3f3f3f3f
 struct Edge {
     int v, nxt, cap, w;
     Edge () {}
     Edge (int v, int nxt, int cap, int w) : v(v), nxt(nxt), cap(cap), w(w) {}
 } edge[N*N];
 int head[N], tot, dis[N], cur[N], pre[N], gap[N], n, m, t;

 void Add(int u, int v, int cap, int w) {
     edge[tot] = Edge(v, head[u], cap, w); head[u] = tot++;
     edge[tot] = Edge(u, head[v], , w); head[v] = tot++;
 }

 int BFS(int S, int T) {
     queue<int> que; que.push(T);
     memset(dis, INF, sizeof(dis));
     memset(gap, , sizeof(gap));
     gap[]++; dis[T] = ;
     while(!que.empty()) {
         int u = que.front(); que.pop();
         for(int i = head[u]; ~i; i = edge[i].nxt) {
             int v = edge[i].v;
             if(dis[v] == INF) {
                 dis[v] = dis[u] + ;
                 gap[dis[v]]++;
                 que.push(v);
             }
         }
     }
 }

 int ISAP(int S, int T, int n) {
     BFS(S, T);
     memcpy(cur, head, sizeof(cur));
     int u = pre[S] = S, i, index, flow, ans = ;
     while(dis[S] < n) {
         if(u == T) {
             flow = INF, index = S; // index = S !!!
             for(i = S; i != T; i = edge[cur[i]].v)
                 if(flow > edge[cur[i]].cap) flow = edge[cur[i]].cap, index = i;
             for(i = S; i != T; i = edge[cur[i]].v)
                 edge[cur[i]].cap -= flow, edge[cur[i]^].cap += flow;
             ans += flow, u = index;
         }
         for(i = cur[u]; ~i; i = edge[i].nxt)
             if(edge[i].cap >  && dis[edge[i].v] == dis[u] - ) break;
         if(~i) {
             pre[edge[i].v] = u; cur[u] = i; u = edge[i].v;
         } else {
             if(--gap[dis[u]] == ) break;
             int md = n;
             for(i = head[u]; ~i; i = edge[i].nxt)
                 if(md > dis[edge[i].v] && edge[i].cap > ) md = dis[edge[i].v], cur[u] = i;
             gap[dis[u] = md + ]++;
             u = pre[u];
         }
     }
     return ans;
 }

 int main() {
     while(~scanf("%d%d%d", &n, &m, &t)) {
         int r = , l = INF;
         memset(head, -, sizeof(head)); tot = ;
         for(int i = ; i <= m; i++) {
             int u, v, w; scanf("%d%d%d", &u, &v, &w);
             Add(u, v, , w);
             if(r < w) r = w;
             if(l > w) l = w;
         }
         int ans = r;
         while(l <= r) {
             int mid = (l + r) >> ;
             for(int i = ; i < tot; i++)
                 if(edge[i].w <= mid) edge[i].cap = ;
                 else edge[i].cap = ;
             if(ISAP(, n, n) >= t) r = mid - , ans = mid;
             else l = mid + ;
         }
         printf("%d\n", ans);
     }
     return ;
 }