Ombrophobic Bovines

Description

FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter.

The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction.

Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse.

Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

Input

* Line 1: Two space-separated integers: F and P

* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i.

* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

Output

* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".

Sample Input

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output

110

Hint

OUTPUT DETAILS:

In 110 time units, two cows from field 1 can get under the shelter in that field, four cows from field 1 can get under the shelter in field 2, and one cow can get to field 3 and join the cows from that field under the shelter in field 3. Although there are other plans that will get all the cows under a shelter, none will do it in fewer than 110 time units.
題意:給出一些牛棚,開始牛棚邊有一些牛,牛棚之間有路相連,走一條路會花費固定的時間。牛在牛棚邊吃草,下雨時牛得躲進牛棚,每個牛棚容量有限。
求:在所有牛都能進牛棚時最少需要多少時間。
分析:如果知道了時間time,則如果兩個牛棚之間所需的最少時間<=time,每個牛棚拆成兩點x1和x2。如果兩個牛棚a和b之間可以相連,則ax1連ax2,容量為無窮(因為路是無限大的,可以容納任意多的牛)。s->x1,容量為每個牛棚容量為牛棚的牛數量,x2->t,容量為每個牛棚的容量。
一定要把每個點拆成兩個點的。如果直接連原圖中的點,是不對的。舉個例子:1->2->3, 邊長度都為1,則1到3是2。如果枚舉ans等于1是,會產生1->2,2->3合法,則最后1也可以到3了,這是錯的。
代碼:
#include <stdio.h>  
#include 
<string.h>  
#include 
<algorithm>
#include 
<iostream>
#define Min(a, b) (a) < (b) ? a : b  
using  namespace std;  
const  int MAXN = 1005;  
const  int MAXM = 210000;  
const  int INF = 1100000000;  
struct  Edge  
{  
    
int  st, ed;  
    
int  next;  
    
int  flow; 
    
int cap; 
}edge[MAXM]; 
int  head[MAXN], level[MAXN], que[MAXN], E;
//const __int64 MAXLEN = 1000000000;
const __int64 MAXLEN = 9999999999999999;
__int64 map[MAXN][MAXN];
int fieldcow[MAXN], fieldcap[MAXN];
void  add(int u, int v, int w)  
{  
    
//printf("add %d %d %d\n", u, v, w);
    edge[E].flow = 0;  
    edge[E].cap 
= w;
    edge[E].st 
= u;  
    edge[E].ed 
= v;  
    edge[E].next 
= head[u];  
    head[u] 
= E++;      
    edge[E].flow 
= 0
    edge[E].cap 
= 0
    edge[E].st 
= v;  
    edge[E].ed 
= u;  
    edge[E].next 
= head[v];  
    head[v] 
= E++;  
}
int  dinic_bfs(int src, int dest, int ver)        
{        
    
int i, j;         
    
for (i = 0; i <= ver; i++)
    {    
        level[i] 
= -1;
    }
    
int rear = 1;        
    que[
0= src; level[src] = 0;        
    
for(i = 0; i < rear; i++
    {        
          
for(j = head[que[i]]; j != -1; j = edge[j].next)
         {        
            
if(level[edge[j].ed] == -1 && edge[j].cap > edge[j].flow)        
            {        
              level[edge[j].ed] 
= level[que[i]]+1;        
              que[rear
++= edge[j].ed;        
            }
         }
    }
    
return  level[dest] >= 0;        
}        
     
int dinic_dfs(int src, int dest, int ver)        
{        
    
int stk[MAXN], top = 0;        
    
int ret = 0, cur, ptr, pre[MAXN], minf, i;        
    
int del[MAXN];        
    
for (i = 0; i <= ver; i++
    {
        del[i] 
= 0;
    }
    stk[top
++= src;         
    pre[src] 
= src; 
    cur 
= src;        
    
while(top)        
    {        
        
while(cur != dest && top)        
        {        
            
for(i = head[cur]; i != -1; i = edge[i].next)        
            {        
                
if(level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow  && !del[edge[i].ed])        
                {        
                    stk[top
++= edge[i].ed;      
                    cur 
= edge[i].ed;        
                    pre[edge[i].ed] 
= i;                       
                    
break;     
                }        
            }     
            
if(i == -1)       
            {        
                del[cur] 
= 1;        
                top
--;        
                
if(top) cur = stk[top-1];        
            }        
        }                
        
if(cur == dest)        
        {       
            minf 
= INF;        
            
while(cur != src)        
            {        
                cur 
= pre[cur];        
                
if(edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;        
                cur 
= edge[cur].st;        
            }
            cur 
= dest;        
            
while(cur != src)        
            {        
                cur 
= pre[cur];        
                edge[cur].flow 
+= minf;        
                edge[cur
^1].flow -= minf;        
                
if(edge[cur].cap - edge[cur].flow == 0)
                {
                     ptr 
= edge[cur].st;
                }
                cur 
= edge[cur].st;        
            }        
            
while(top > 0&& stk[top-1!= ptr) top--;        
            
if(top)  cur = stk[top-1];        
            ret 
+= minf;      
        }        
    }        
    
return ret;        
}        
int Dinic(int src, int dest, int ver)        
{        
    
int  ret = 0, t;        
    
while(dinic_bfs(src, dest, ver))        
    {        
        t 
= dinic_dfs(src, dest, ver);        
        
if(t) ret += t;        
        
else  break;        
    }        
    
return ret;        
}
void floyd(int n)
{
    
int k, i, j;
    
for (k = 1; k <= n; k++)
    {
        
for (i = 1; i <= n; i++)
        {
            
for (j = 1; j <= n; j++)
            {
                
//可以自己到自己的。因為自己對自己可容納牛。 
                if (map[i][k] != MAXLEN && map[k][j] != MAXLEN && map[i][k] + map[k][j] < map[i][j])
                {
                    map[i][j] 
= map[i][k] + map[k][j];
                }
            }
        }
    }
}
void build(int n, __int64 distance)
{
    
int s = 0, t = n + n + 1, ver = t + 1, i, j;
    E 
= 0;
    
for (i = 0; i <= ver; i++)
    {
        head[i] 
= -1;
    }
    
for (i = 1; i <= n; i++)
    {
        
for (j = 1; j <= n; j++)
        {
            
if (map[i][j] <= distance)
            {
                add(i, j 
+ n, INF);
            }
        }
    }
    
for (i = 1; i <= n; i++)
    {
        add(
0, i, fieldcow[i]);
        add(i 
+ n, t, fieldcap[i]);
    }
}
int main()
{
    
int n, m, i, cownum, j, u, v;
    __int64 w, limit 
= 0;
    scanf(
"%d%d"&n, &m);
    
for (i = 1, cownum = 0; i <= n; i++)
    {
        scanf(
"%d%d"&fieldcow[i], &fieldcap[i]);
        cownum 
+= fieldcow[i];
    }
    
for (i = 1; i <= n; i++)
    {
        
for (j = 1; j <= n; j++)
        {
            map[i][j] 
= MAXLEN;            
        }
        map[i][i] 
= 0;
    }
    
while (m--)
    {
        scanf(
"%d%d%I64d"&u, &v, &w);
        limit 
+= w;
        
if (map[u][v] > w)
        {
            map[u][v] 
= map[v][u] = w;
        }
    }
    floyd(n);
    
int s = 0, t = n + n + 1, ver = t + 1, ans, sign = 0;
    __int64 l 
= 0, r = limit + 1, mid;
    
while (l < r)
    {
        mid 
= (l + r) >> 1;
         build(n, mid);
         ans 
= Dinic(0, t, ver);
         
if (ans == cownum)
         {
             r 
= mid;
             sign 
= 1;
         }
         
else
         {
             l 
= mid + 1;
         }
    }
    
if (sign)//如果沒有滿足所有牛都進牛棚會輸出-1 
    {
        printf(
"%I64d\n", r);
    }
    
else
    {
        printf(
"-1\n");
    }
    
return 0;
}
/*
2 1
1 2
2 0
1 2 10

3 1
1 2
1 0
3 3
1 2 10

3 1
1 2
1 0
3 2
1 2 10

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

1 0
1 2

1 0
2 1
*/