HDU 4280 Island Transport

Island Transport

Time Limit: 10000ms

Memory Limit: 65536KB

This problem will be judged on HDU. Original ID: 4280
64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

9

6

Source

2012 ACM/ICPC Asia Regional Tianjin Online

解题：网络流模板题，注意双向边，其实可以这样子搞~

 #include <bits/stdc++.h>

 using namespace std;

 const int INF = ~0U>>;

 const int maxn = ;

 struct arc{

     int to,flow,next;

     arc(int x = ,int y = ,int z = -){

         to = x;

         flow = y;

         next = z;

     }

 }e[];

 int head[maxn],d[maxn],gap[maxn],tot,S,T,n,m;

 void add(int u,int v,int flow){

     e[tot] = arc(v,flow,head[u]);

     head[u] = tot++;

     e[tot] = arc(u,flow,head[v]);

     head[v] = tot++;

 }

 queue<int>q;

 void bfs(){

     for(int i = ; i <= n; ++i){

         d[i] = -;

         gap[i] = ;

     }

     d[T] = ;

     q.push(T);

     while(!q.empty()){

         int u = q.front();

         q.pop();

         ++gap[d[u]];

         for(int i = head[u]; ~i; i = e[i].next){

             if(d[e[i].to] == -){

                 d[e[i].to] = d[u] + ;

                 q.push(e[i].to);

             }

         }

     }

 }

 int dfs(int u,int low){

     if(u == T) return low;

     int tmp = ,minH = n - ;

     for(int i = head[u]; ~i; i = e[i].next){

         if(e[i].flow){

             if(d[e[i].to] +  == d[u]){

                 int a = dfs(e[i].to,min(low,e[i].flow));

                 e[i].flow -= a;

                 e[i^].flow += a;

                 low -= a;

                 tmp += a;

                 if(!low) break;

                 if(d[S] >= n) return tmp;

             }

             if(e[i].flow) minH = min(minH,d[e[i].to]);

         }

     }

     if(!tmp){

         if(--gap[d[u]] == ) d[S] = n;

         ++gap[d[u] = minH + ];

     }

     return tmp;

 }

 int main(){

     int kase,x,y,s,t,u,v,w;

     scanf("%d",&kase);

     while(kase--){

         scanf("%d%d",&n,&m);

         memset(head,-,sizeof head);

         s = INF;

         int ret = t = ;

         for(int i = ; i <= n; ++i){

             scanf("%d%d",&x,&y);

             if(s > x){

                 s = x;

                 S = i;

             }

             if(t < x){

                 t = x;

                 T = i;

             }

         }

         for(int i = tot = ; i < m; ++i){

             scanf("%d%d%d",&u,&v,&w);

             add(u,v,w);

         }

         bfs();

         while(d[S] < n) ret += dfs(S,INF);

         printf("%d\n",ret);

     }

     return ;

 }

巴特西