HDU 4280 Island Transport(网络流,最大流)

HDU 4280 Island Transport(网络流,最大流)

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

Http

HDU:https://vjudge.net/problem/HDU-4280

Source

网络流,最大流

题目大意

在n个岛屿中,有m条双向航线,航线有单位时间内的运输上限,现在求从最左侧到最右侧的最大运输量

解决思路

这道题很明显是最大流,但数据范围有点大,似乎是专门卡Dinic算法的,可以当做是一道联系Dinic卡时的好题。反正能减的就减。
如果不知道Dinic最大流算法,可以到我的这篇文章查看

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
//像这种卡时的题目尽量不要用STL
#define min(x,y) x>y? y : x
#define RG register

const int maxN=200031;//数组一定要开够
const int maxM=500031;
const int inf=2147483647;

struct Edge
{
    int v,flow;
};

int n,m;
int S,T;
int cnt=-1;
int Head[maxN];
int Next[maxM];
Edge E[maxM];
int cur[maxN];
int depth[maxN];
int Queue[maxN];

inline void Add_Edge(int u,int v,int flow);
int dfs(int u,int flow);
inline int read();

int main()
{
    int Case;
    Case=read();
    while (Case--)//多组数据
    {
        memset(Head,-1,sizeof(Head));//注意这里去掉了对Next的初始化,因为我们在Add_Edge函数中已经考虑到了,不需要初始化
        cnt=-1;
        n=read();
        m=read();
        int minx=inf,maxx=-inf;
        for (int i=1;i<=n;i++)
        {
            int x=read(),y=read();
            if (x<minx)
            {
                minx=x;
                S=i;
            }
            if (x>maxx)
            {
                maxx=x;
                T=i;
            }
        }
        for (int i=1;i<=m;i++)
        {
            int u=read(),v=read(),flow=read();
            Add_Edge(u,v,flow);//注意,因为是双向边,我们不要像原来那样建流量为0的反边(那样就有4条边了),只要建两条流量均为flow的边
        }
        int Ans=0;
        while (1)
        {
            memset(depth,-1,sizeof(depth));//减少函数调用,把bfs放入主过程
            int Top=1,Tail=0;
            depth[S]=1;
            Queue[1]=S;
            do
            {
                Tail++;
                int u=Queue[Tail];
                if (u==T)
                    break;
                for (int i=Head[u];i!=-1;i=Next[i])
                {
                    int v=E[i].v;
                    if ((E[i].flow>0)&&(depth[v]==-1))
                    {
                        depth[v]=depth[u]+1;
                        Top++;
                        Queue[Top]=v;
                    }
                }
            }
            while (Top!=Tail);
            if (depth[T]==-1)
                break;
            for (int i=1;i<=n;i++)//当前弧优化,这个优化力度很明显,一定要加
                cur[i]=Head[i];
            
            while (int di=dfs(S,inf))//dfs寻找增广路
                Ans+=di;
        }
        printf("%d\n",Ans);
    }
    return 0;
}

inline int read()//读入优化,不知道为什么scanf比读入优化快300ms,可能是评测机的问题
{
    int x=0;
    scanf("%d",&x);//这里直接用scanf读入
    return x;
    int k=1;
    char ch=getchar();
    while (((ch>'9')||(ch<'0'))&&(ch!='-'))
        ch=getchar();
    if (ch=='-')
    {
        ch=getchar();
        k=-1;
    }
    while ((ch>='0')&&(ch<='9'))
    {
        x=x*10+ch-48;
        ch=getchar();
    }
    return x*k;
}

inline void Add_Edge(int u,int v,int flow)//添加边,注意正反向都是流量为flow,因为是双向边
{
    cnt++;
    Next[cnt]=Head[u];
    Head[u]=cnt;
    E[cnt].v=v;
    E[cnt].flow=flow;

    cnt++;
    Next[cnt]=Head[v];
    Head[v]=cnt;
    E[cnt].v=u;
    E[cnt].flow=flow;
    return;
}

int dfs(int u,int flow)//dfs寻增广路
{
    if (u==T)
        return flow;
    for (int &i=cur[u];i!=-1;i=Next[i])
    {
        int v=E[i].v;
        if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
        {
            int di=dfs(v,min(flow,E[i].flow));
            if (di>0)
            {
                E[i].flow-=di;
                E[i^1].flow+=di;
                return di;
            }
        }
    }
    return 0;
}

另附上我的TLE霸屏截图:
此处输入图片的描述
中间两次Mayuri是我用小号交的网上题解的代码,时间也是卡在超时边缘。
前三次AC分别是:scanf读入,去掉一些循环内的冗余部分,原AC代码
第二次AC与第一次AC中间的TLE是去掉当前弧优化后的代码。
代码均已开源(vjudge),可以去看一看是如何一步一步卡时的。

posted @ 2017-08-01 08:05 SYCstudio 阅读(...) 评论(...) 编辑 收藏