题意：
有n个点，每个点有两个权值，金子数量还有仓库容量，金子可以存在自己的仓库里或者是别的仓库里，仓库和仓库之间有距离，问所有金子都必须存到库里最大距离的最小是多少？
思路：
比较简单，方法也不唯一，大体可以这样，先二分，然后用最大流或者匹配..来判断是不是满足就行了，我用的是二分最大流，具体代码在下面。
#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N_node 400 + 50
#define N_edge 200 * 200 * 2 + 10000
#define INF 1000000000
using namespace std;
typedef struct
{
int to ,cost ,next;
}STAR;
typedef struct
{
int a ,b ,c;
}EDGE;
typedef struct
{
int x ,t;
}DEP;
STAR E[N_edge];
EDGE edge[N_edge];
DEP xin ,tou;
int list[N_node] ,listt[N_node] ,tot;
int deep[N_node];
int c1[N_node] ,c2[N_node];
int num[N_edge] ,numt[N_edge];
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;
queue<DEP>q;
q.push(xin);
deep[s] = 0;
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 ++)
listt[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 = listt[s] ;k ;k = E[k].next)
{
listt[s] = k;
int c = E[k].cost ,to = E[k].to;
if(!c || deep[to] != deep[s] + 1)
continue;
int tmp = DFS_Flow(to ,t ,minn(c ,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;
}
int GetMaxFlow(int mid ,int n ,int m)
{
memset(list ,0 ,sizeof(list));
tot = 1;
for(int i = 1 ;i <= n ;i ++)
{
add(0 ,i ,c1[i]);
add(i ,i + n ,INF);
add(i + n ,n + n + 1 ,c2[i]);
}
for(int i = 1 ;i <= m ;i ++)
{
if(edge[i].c <= mid)
{
add(edge[i].a ,edge[i].b + n ,INF);
add(edge[i].b ,edge[i].a + n ,INF);
}
}
return DINIC(0 ,n + n + 1 ,n + n + 1);
}
int main ()
{
int n ,m ,i ,s1 ,s2;
while(~scanf("%d" ,&n) && n)
{
s1 = s2 = 0;
for(i = 1 ;i <= n ;i ++)
{
scanf("%d" ,&c1[i]);
s1 += c1[i];
}
for(i = 1 ;i <= n ;i ++)
{
scanf("%d" ,&c2[i]);
s2 += c2[i];
}
scanf("%d" ,&m);
for(i = 1 ;i <= m ;i ++)
{
scanf("%d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c);
numt[i] = edge[i].c;
}
if(s2 < s1)
{
printf("No Solution\n");
continue;
}
sort(numt + 1 ,numt + m + 1);
int numid = 0;
for(i = 1 ;i <= m ;i ++)
if(i == 1 || numt[i] != numt[i-1])
num[++numid] = numt[i];
num[0] = 0;
int low = 0 ,up = numid ,mid ,ans = -1;
while(low <= up)
{
mid = (low + up) >> 1;
if(GetMaxFlow(num[mid] ,n ,m) == s1)
{
ans = num[mid];
up = mid - 1;
}
else low = mid + 1;
}
if(ans == -1) printf("No Solution\n");
else printf("%d\n" ,ans);
}
return 0;
}
