Island Transport
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10716 Accepted Submission(s): 3430
Problem Description
In
the vast waters far far away, there are many islands. People are living
on the islands, and all the transport among the islands relies on the
ships.
You have a transportation company there. Some routes are
opened for passengers. Each route is a straight line connecting two
different islands, and it is bidirectional. Within an hour, a route can
transport a certain number of passengers in one direction. For safety,
no two routes are cross or overlap and no routes will pass an island
except the departing island and the arriving island. Each island can be
treated as a point on the XY plane coordinate system. X coordinate
increase from west to east, and Y coordinate increase from south to
north.
The transport capacity is important to you. Suppose many
passengers depart from the westernmost island and would like to arrive
at the easternmost island, the maximum number of passengers arrive at
the latter within every hour is the transport capacity. Please calculate
it.
Input
The first line contains one integer T (1<=T<=20), the number of test cases.
Then
T test cases follow. The first line of each test case contains two
integers N and M (2<=N,M<=100000), the number of islands and the
number of routes. Islands are number from 1 to N.
Then N lines
follow. Each line contain two integers, the X and Y coordinate of an
island. The K-th line in the N lines describes the island K. The
absolute values of all the coordinates are no more than 100000.
Then
M lines follow. Each line contains three integers I1, I2
(1<=I1,I2<=N) and C (1<=C<=10000) . It means there is a
route connecting island I1 and island I2, and it can transport C
passengers in one direction within an hour.
It is guaranteed that
the routes obey the rules described above. There is only one island is
westernmost and only one island is easternmost. No two islands would
have the same coordinates. Each island can go to any other island by the
routes.
Output
For each test case, output an integer in one line, the transport capacity.
Sample Input
2
5 7
3 3
3 0
3 1
0 0
4 5
1 3 3
2 3 4
2 4 3
1 5 6
4 5 3
1 4 4
3 4 2
6 7
-1 -1
0 1
0 2
1 0
1 1
2 3
1 2 1
2 3 6
4 5 5
5 6 3
1 4 6
2 5 5
3 6 4
Sample Output
9
6
分析：比较裸的最大流，先找起点(X坐标最小)和终点(X坐标最大),
用数组模拟邻接链表记录图的边，跑一遍Dinic算出最大流。
用G++提交8330MS,用C++提交Time Limit Exceeded......
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define INF 1000000007
using namespace std;
int N,M,cnt,T,S;//T是终点
struct Node{
int t,c;
}edge[500000];
int head[100010],next1[500000];
int dis[100010];
void add(int s,int t,int c)
{
edge[cnt].t=t;
edge[cnt].c=c;
next1[cnt]=head[s];
head[s]=cnt++;
}
int bfs()
{
memset(dis,-1,sizeof(dis));
dis[S]=0;
queue<int> q;
q.push(S);
while(!q.empty())
{
int u=q.front();
q.pop();
int k=head[u];
while(k!=-1)
{
int v=edge[k].t;
if(dis[v]==-1&&edge[k].c>0)
{
dis[v]=dis[u]+1;
q.push(v);
}
k=next1[k];
}
}
if(dis[T]>0) return 1;
return 0;
}
int dfs(int cur,int m)
{
if(cur==T) return m;
int res=0,f,k=head[cur];
while(k!=-1)
{
if(dis[edge[k].t]==dis[cur]+1&&edge[k].c>0&&(f=dfs(edge[k].t,min(m,edge[k].c))))
{
edge[k].c-=f;
edge[k^1].c+=f;
res+=f;
m-=f;
if(!m) break;
}
k=next1[k];
}
if(!res) dis[cur]=-2;
return res;
}
int main()
{
int cas;
scanf("%d",&cas);
while(cas--)
{
memset(head,-1,sizeof(head));
scanf("%d%d",&N,&M);
int sx=INF,tx=-INF;//起点S和终点T的x坐标
cnt=S=T=0;
for(int i=1;i<=N;i++)
{
int x,y;
scanf("%d%d",&x,&y);
if(sx>=x) {sx=x;S=i;}
if(tx<=x) {tx=x;T=i;}
}
for(int i=0;i<M;i++)
{
int s,t,c;
scanf("%d%d%d",&s,&t,&c);
add(s,t,c);
add(t,s,c);
}
int ans=0,res;
while(bfs())
while(res=dfs(S,INF)) ans+=res;
printf("%d\n",ans);
}
return 0;
}
原文地址：https://www.cnblogs.com/ACRykl/p/8859076.html