学步园

题意：有N个地点，FJ 想从1走到N 每条边只能走一遍 走T次 求在满足条件下，最大的边最小。
思路：这题跟 poj 3228 Gold Transportation(二分+最大流)
这题差不多，就是用二分求出满足条件下的最小边权 EK超时，此题用sap做的

//3212K   
797MS
#include <stdio.h>
#include <string.h>
#define VM 210
#define EM
160010     
//建边重复1遍，所以是4*40000

int head[VM],oldhead[VM],dep[VM],gap[VM],cur[VM],pre[VM];
int e1,e,n,p,t,src,des;
struct E
{
    int
to,cap,nxt;
}edge[EM],oldedge[EM];

void addedge1 (int cu,int cv,int cw)
{
   
oldedge[e1].to = cv;
   
oldedge[e1].cap = cw;
   
oldedge[e1].nxt = oldhead[cu];
    oldhead[cu]
= e1 ++;
}
void addedge (int cu,int cv,int cw)
{
    edge[e].to =
cv;
    edge[e].cap
= cw;
    edge[e].nxt
= head[cu];
    head[cu] = e
++;
}

void Build (int num)  //构建一新的图
{
    memset
(head,-1,sizeof(head));
    e = 0;
    for (int u =
1;u <= n;u ++)
       
for (int i = oldhead[u];i != -1;i = oldedge[i].nxt)
       
{
           
int v = oldedge[i].to;
           
if (oldedge[i].cap <= num)
           
{
               
addedge
(u,v,1);   
//边权为1就可以了
               
addedge (v,u,1);
           
}
       
}
}

int Sap ()   //sap算法，不解释
{
    memset
(dep,0,sizeof(dep));
    memset
(gap,0,sizeof(gap));
    memcpy
(cur,head,sizeof(head));
    int u =
pre[src] = src;
    int res =
0;
    gap[0] =
n;
    while
(dep[src] < n)
    {
    loop:
       
for (int &i = cur[u];i != -1;i = edge[i].nxt)
       
{
           
int v = edge[i].to;
           
if (edge[i].cap && dep[u] == dep[v]
+ 1)
           
{
               
pre[v] = u;