# 【bzoj1738】[Usaco2005 mar]Ombrophobic Bovines 发抖的牛 Floyd+二分+网络流最大流

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.

* 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.

第1行：两个整数F和P;
第2到F+1行：第i+l行有两个整数描述第i个田地，第一个表示田地上的牛数，第二个表示田地上的雨棚容量．两个整数都在0和1000之间．
第F+2到F+P+I行：每行三个整数描述一条路，分别是起点终点，及通过这条路所需的时间（在1和10^9之间）．

* 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".

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

110

floyd+二分+拆点+网络流

#include <cstdio>
#include <cstring>
#include <queue>
#define inf 0x3fffffff
using namespace std;
queue<int> q;
long long dis[201][201];
int a[201] , b[201] , head[403] , to[180000] , val[180000] , next[180000] , cnt , s , t , deep[403];
void add(int x , int y , long long z)
{
to[++cnt] = y;
val[cnt] = z;
}
bool bfs()
{
int x , i;
while(!q.empty())
q.pop();
memset(deep , 0 , sizeof(deep));
deep[s] = 1;
q.push(s);
while(!q.empty())
{
x = q.front();
q.pop();
for(i = head[x] ; i ; i = next[i])
{
if(val[i] && !deep[to[i]])
{
deep[to[i]] = deep[x] + 1;
if(to[i] == t)
return 1;
q.push(to[i]);
}
}
}
return 0;
}
int dinic(int x , int low)
{
if(x == t)
return low;
int temp = low , i , k;
for(i = head[x] ; i ; i = next[i])
{
if(val[i] && deep[to[i]] == deep[x] + 1)
{
k = dinic(to[i] , min(temp , val[i]));
if(!k) deep[to[i]] = 0;
val[i] -= k;
val[i ^ 1] += k;
if(!(temp -= k)) break;
}
}
return low - temp;
}
bool judge(int n , long long mid , int sum)
{
memset(to , 0 , sizeof(to));
memset(val , 0 , sizeof(val));
memset(next , 0 , sizeof(next));
cnt = 1;
int i , j , maxflow = 0;
for(i = 1 ; i <= n ; i ++ )
{
add(s , i , a[i]);
add(i , s , 0);
add(i + n , t , b[i]);
add(t , i + n , 0);
for(j = 1 ; j <= n ; j ++ )
{
if(i == j || dis[i][j] <= mid)
add(i , j + n , inf) , add(j + n , i , 0);
}
}
while(bfs())
maxflow += dinic(s , inf);
return maxflow == sum;
}
int main()
{
int n , m , i , j , k , x , y , suma = 0 , sumb = 0;
long long z , l = 0 , r = 0 , mid , ans = -1;
scanf("%d%d" , &n , &m);
s = 0 , t = 2 * n + 1;
for(i = 1 ; i <= n ; i ++ )
scanf("%d%d" , &a[i] , &b[i]) , suma += a[i] , sumb += b[i];
memset(dis , 0x3f , sizeof(dis));
for(i = 1 ; i <= m ; i ++ )
scanf("%d%d%lld" , &x , &y , &z) , dis[x][y] = dis[y][x] = min(dis[x][y] , z);
if(suma > sumb)
{
printf("-1\n");
return 0;
}
for(k = 1 ; k <= n ; k ++ )
for(i = 1 ; i <= n ; i ++ )
for(j = 1 ; j <= n ; j ++ )
dis[i][j] = min(dis[i][j] , dis[i][k] + dis[k][j]);
for(i = 1 ; i <= n ; i ++ )
for(j = 1 ; j <= n ; j ++ )
if(i != j)
r = max(r , dis[i][j]);
while(l <= r)
{
mid = (l + r) >> 1;
if(judge(n , mid , suma))
ans = mid , r = mid - 1;
else
l = mid + 1;
}
printf("%lld\n" , ans < 10000000000000ll ? ans : -1);
return 0;
}
posted @ 2017-02-13 16:28  GXZlegend  阅读(300)  评论(0编辑  收藏