解题报告 之 POJ2391 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
题目大意：有n个棚，有m条路，每个棚紧挨着一些牛，每个棚有一定容量。每条路有一个长度，每条路可以同时跑无数条牛。问要让所有牛有棚住，牛中移动距离的最长路最小能有多小？
分析：与前面的题类似，先跑Floyd，然后都是二分，然后跑最大流，直到满足为止。但是这个题建图有一个很困难的地方。就是拆点及拆点的作用。这里把每个棚拆成两个点，负载为INF，关键在于如果两个点u->v可达且路径长度<=mid，那么此时在u的入点连一条边到v的出点。为什么那么奇葩呢，因为转移的量只能转移一次，如果不拆点流量可能转移多次，那么总路径长度将会大于mid。好好体会一下！
另外不知道为什么二分用 while(low<high){ if(...) low=mid+1;else high=mid} 输出low不行，而用ans记录法（见代码）就可以，求大神解答，叩谢！
上代码
<span style="font-size:18px;">#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int MAXN = 510;
const int MAXM = 330110;
const int inf = 0x3f3f3f3f;
const long long INF = 1e16;
struct Edge
{
int to, cap, next;
};
Edge edge[MAXM];
int level[MAXN];
int head[MAXN];
int cow[MAXN];
int hold[MAXN];
long long dist[MAXN][MAXN];
int src, des, cnt;
void addedge( int from, int to, int cap )
{
edge[cnt].to = to;
edge[cnt].cap = cap;
edge[cnt].next = head[from];
head[from] = cnt++;
edge[cnt].to = from;
edge[cnt].cap = 0;
edge[cnt].next = head[to];
head[to] = cnt++;
}
int bfs( )
{
queue<int> q;
while (!q.empty( ))
q.pop( );
memset( level, -1, sizeof level );
level[src] = 0;
q.push( src );
while (!q.empty( ))
{
int u = q.front( );
q.pop( );
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > 0 && level[v] == -1)
{
level[v] = level[u] + 1;
q.push( v );
}
}
}
return level[des] != -1;
}
int dfs( int u, int f )
{
if (u == des) return f;
int tem;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > 0 && level[v] == level[u] + 1)
{
tem = dfs( v, min( f, edge[i].cap ) );
if (tem > 0)
{
edge[i].cap -= tem;
edge[i ^ 1].cap += tem;
return tem;
}
}
}
level[u] = -1;
return 0;
}
int Dinic( )
{
int ans = 0, tem;
while (bfs( ))
{
while (tem = dfs( src, INF ))
{
ans += tem;
}
}
return ans;
}
int main( )
{
int n, m;
while (cin >> n >> m)
{
int cows = 0;
memset( dist, -1, sizeof dist );
src = 0, des = 501;
for (int i = 1; i <= n; i++)
{
scanf( "%d%d", &cow[i], &hold[i] );
cows += cow[i];
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
dist[i][j] = INF;
}
for (int i = 1; i <= m; i++)
{
int a, b, dis;
scanf( "%d%d%d", &a, &b, &dis );
if (dis < dist[a][b])
dist[a][b] = dist[b][a] = dis;
}
//Floyd
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (dist[i][j] > dist[i][k] + dist[k][j])
{
dist[i][j] = dist[i][k] + dist[k][j];
}
long long low = -1, high = INF - 1;
long long ans = -1;
while (low <= high)
{
long long mid = (low + high) / 2;
memset( head, -1, sizeof head );
cnt = 0;
for (int i = 1; i <= n; i++)
{
addedge( src, i, cow[i] );
addedge( i + 250, des, hold[i] );
addedge( i, i + 250, hold[i] );
}
for (int i = 1; i <= n; i++)
{
for (int j = i + 1; j <= n; j++)
{
if (dist[i][j] <= mid)
{
addedge( i, j + 250, inf );
addedge( j, i + 250, inf );
}
}
}
if (Dinic( ) ==cows)
{
ans = mid;
high = mid-1;
}
else low = mid+1;
}
cout << ans << endl;
}
return 0;
}</span>
今天累了，，洗洗睡！