Island Transport

Island Transport

http://acm.hdu.edu.cn/showproblem.php?pid=4280

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 13032    Accepted Submission(s): 4097

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

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

 

因为是双向的，所以add函数里面要改动下

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define maxn 100005
#define MAXN 100005
#define mem(a,b) memset(a,b,sizeof(a))
const int N=200005;
const int M=200005;
const int INF=0x3f3f3f3f;
using namespace std;
int n;
struct Edge{
    int v,next;
    int cap,flow;
}edge[MAXN*20];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
int cnt=0;//实际存储总边数
void isap_init()
{
    cnt=0;
    memset(pre,-1,sizeof(pre));
}
void isap_add(int u,int v,int w)//加边
{
    edge[cnt].v=v;
    edge[cnt].cap=w;
    edge[cnt].flow=0;
    edge[cnt].next=pre[u];
    pre[u]=cnt++;
}
void add(int u,int v,int w){
    isap_add(u,v,w);
    isap_add(v,u,w);
}
bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    dep[t]=0;
    queue<int>q;
    while(!q.empty())
    q.pop();
    q.push(t);//从汇点开始反向建层次图
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是从汇点反向bfs，但应该判断正向弧的余量
            {
                dep[v]=dep[u]+1;
                gap[dep[v]]++;
                q.push(v);
                //if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
                //break;
            }
        }
    }
    return dep[s]!=-1;
}
int isap(int s,int t)
{
    if(!bfs(s,t))
    return 0;
    memcpy(cur,pre,sizeof(pre));
    //for(int i=1;i<=n;i++)
    //cout<<"cur "<<cur[i]<<endl;
    int u=s;
    path[u]=-1;
    int ans=0;
    while(dep[s]<n)//迭代寻找增广路,n为节点数
    {
        if(u==t)
        {
            int f=INF;
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路
                f=min(f,edge[i].cap-edge[i].flow);
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])
            {
                edge[i].flow+=f;
                edge[i^1].flow-=f;
            }
            ans+=f;
            u=s;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
            {
                cur[u]=path[v]=i;//当前弧优化
                flag=true;
                break;
            }
        }
        if(flag)
        {
            u=v;
            continue;
        }
        int x=n;
        if(!(--gap[dep[u]]))return ans;//gap优化
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
            {
                x=dep[edge[i].v];
                cur[u]=i;//常数优化
            }
        }
        dep[u]=x+1;
        gap[dep[u]]++;
        if(u!=s)//当前点没有增广路则后退一个点
        u=edge[path[u]^1].v;
     }
     return ans;
}

struct Point{
    int x,y;
}p[maxn];

int main(){
    std::ios::sync_with_stdio(false);
    int m,s,t;
    int T;
    cin>>T;
    while(T--){
        cin>>n>>m;
        for(int i=1;i<=n;i++) cin>>p[i].x>>p[i].y;
        int a,b,c;
        isap_init();
        for(int i=1;i<=m;i++){
            cin>>a>>b>>c;
            add(a,b,c);
        }
        Point tmp=p[1];
        s=1;
        for(int i=1;i<=n;i++){
            if(tmp.x>p[i].x){
                tmp=p[i];
                s=i;
            }
        }
        tmp=p[1];
        t=1;
        for(int i=1;i<=n;i++){
            if(tmp.x<p[i].x){
                tmp=p[i];
                t=i;
            }
        }
        cout<<isap(s,t)<<endl;
    }
}