题意：
给你N个位置，每个位置都有金矿数量和仓库数量，然后位置和位置之间的距离给了出来，最后问你吧所有的金矿都放到库里面走的路径 最长的最短 是多少？
思路：
比较简单的一个题，直接二分答案，然后用最大流是否满流来判断二分方向，还有就是建图的时候不用拆点什么的，一开始建图想麻烦了，都快敲完了才反应过来，具体看代码。
#include<stdio.h>
#include<string.h>
#include<queue>
#define N_node 200 + 10
#define N_edge 90000
#define INF 1000000000
using namespace std;
typedef struct
{
int to ,cost ,next;
}STAR;
typedef struct
{
int x ,t;
}DEP;
typedef struct
{
int a ,b ,c;
}EDGE;
EDGE edge[22000];
STAR E[N_edge];
DEP xin ,tou;
int list[N_node] ,list2[N_node] ,tot;
int deep[N_node];
int aaa[220] ,bbb[220];
void add(int a ,int b ,int c)
{
E[++tot].to = b;
E[tot].cost = c;
E[tot].next = list[a];
list[a] = tot;
E[++tot].to = a;
E[tot].cost = c;
E[tot].next = list[b];
list[b] = tot;
}
bool BFS_Deep(int s ,int t ,int n)
{
memset(deep ,255 ,sizeof(deep));
xin.x = s ,xin.t = 0;
deep[xin.x] = xin.t;
queue<DEP>q;
q.push(xin);
while(!q.empty())
{
tou = q.front();
q.pop();
for(int k = list[tou.x] ;k ;k = E[k].next)
{
xin.x = E[k].to;
xin.t = tou.t + 1;
if(deep[xin.x] != -1 || !E[k].cost)
continue;
deep[xin.x] = xin.t;
q.push(xin);
}
}
for(int i = 0 ;i <= n ;i ++)
list2[i] = list[i];
return deep[t] != -1;
}
int minn(int x ,int y)
{
return x < y ? x : y;
}
int DFS_Flow(int s ,int t ,int flow)
{
if(s == t) return flow;
int nowflow = 0;
for(int k = list2[s] ;k ;k = E[k].next)
{
list2[s] = k;
int to = E[k].to;
if(deep[to] != deep[s] + 1 || !E[k].cost)
continue;
int tmp = DFS_Flow(to ,t ,minn(E[k].cost ,flow - nowflow));
nowflow += tmp;
E[k].cost -= tmp;
E[k^1].cost += tmp;
if(nowflow == flow) break;
}
if(!nowflow) deep[s] = 0;
return nowflow;
}
int DINIC(int s ,int t ,int n)
{
int Ans = 0;
while(BFS_Deep(s ,t ,n))
{
Ans += DFS_Flow(s ,t ,INF);
}
return Ans;
}
bool ok(int n ,int m ,int sum ,int mid)
{
memset(list ,0 ,sizeof(list)) ,tot = 1;
for(int i = 1 ;i <= n ;i ++)
{
add(0 ,i ,aaa[i]);
add(i ,n + 1 ,bbb[i]);
}
for(int i = 1 ;i <= m ;i ++)
if(edge[i].c <= mid)
{
add(edge[i].a ,edge[i].b ,INF);
}
int flow = DINIC(0 ,n + 1 ,n + 1);
return flow == sum;
}
int main ()
{
int n ,m ,i ,sum;
while(~scanf("%d" ,&n) && n)
{
for(sum = 0 ,i = 1 ;i <= n ;i ++)
{
scanf("%d" ,&aaa[i]);
sum += aaa[i];
}
for(i = 1 ;i <= n ;i ++)
scanf("%d" ,&bbb[i]);
scanf("%d" ,&m);
for(i = 1 ;i <= m ;i ++)
scanf("%d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c);
int low ,up ,mid ,Ans = -1;
low = 0 ,up = 11000;
while(low <= up)
{
mid = (low + up) >> 1;
if(ok(n ,m ,sum ,mid))
{
Ans = mid;
up = mid - 1;
}
else low = mid + 1;
}
Ans == -1 ? puts("No Solution"):printf("%d\n" ,Ans);
}
return 0;
}
