题目描述
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.
输入
* 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.
输出
* 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".
样例输入
3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120
样例输出
110
题解
floyd+二分+拆点+网络流
先用floyd求出任意两点之间的距离。
然后二分答案，若i与j之间的距离小于等于mid，则将i与j'（拆出来的点）间连一条容量为正无穷的边。
将源点与每个点间连一条容量为牛数的边，将每个拆出来的点与汇点间连一条容量为牛棚容量的边。
然后跑网络流，判断是否满流即可。
注意图可以是不连通的，所以当ans过大时，说明必须要用到题目中不存在的边，即无论如何都不能满足题意，输出-1。
注意距离要开long long。
#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;
next[cnt] = head[x];
head[x] = cnt;
}
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(head , 0 , sizeof(head));
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
");
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
" , ans < 10000000000000ll ? ans : -1);
return 0;
}