HDU4280(Island Transport )最大流SAP算法+当前弧优化

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 

/*
 *题目大意:
 *有N个岛屿,M条无向路,每个路有一最大允许的客流量,求从最西的那个岛屿最多能运用多少乘客到最东的那个岛屿;
 *
 *算法思想:
 *最大流SAP算法+当前弧优化
 *当前弧优化
 *为了使每次找增广路的时间变成均摊O(V),对于每个点保存“当前弧”:初始时当前弧是邻接表的第一条弧;
 *在邻接表中查找时从当前弧开始查找,找到了一条允许弧,就把这条弧设为当前弧;
 *改变距离标号时,把当前弧重新设为邻接表的第一条弧;
 *
**/
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int N=100010;
const int M=400010;
const int INF=0xfffffff;

int n,m,cnt;

struct Edge
{
    int v , cap , next;
} edge[M];

int head[N],pre[N],d[N],numd[N];//分别为链表的头指针，每个点的前驱，每个点的d值，以及标号为d[i] 的点的个数
int cur_edge[N];//从每个点出发满足d[i] = d[j] + 1的边的地址 ， 插入边时的计数，源点与汇点

void AddEdge(int u,int v,int c)
{
    edge[cnt].v=v;
    edge[cnt].cap=c;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cap=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}

void bfs(int sink)//先用广度优先算出每个点的d值
{
    memset(numd,0,sizeof(numd));
    for(int i=1; i<=n; i++)
        numd[d[i]=n]++;
    d[sink]=0;
    numd[n]--;
    numd[0]++;
    queue Q;
    Q.push(sink);

    while(!Q.empty())
    {
        int v=Q.front();
        Q.pop();

        int i=head[v];
        while(i!=-1)
        {
            int u=edge[i].v;

            if(d[u]edge[cur_edge[from]].cap)
                {
                    neck=from;
                    cur_flow=edge[cur_edge[from]].cap;
                }
            }

            for(int from=source; from!=sink; from=edge[cur_edge[from]].v) //修改增广路上的边的容量
            {
                int tmp=cur_edge[from];
                edge[tmp].cap-=cur_flow;
                edge[tmp^1].cap+=cur_flow;
            }
            max_flow+=cur_flow;//累加计算最大流
            u=neck;//下一次搜索直接从瓶颈边的前一个节点搜起
        }

        int i;
        for(i=cur_edge[u]; i!=-1; i=edge[i].next) //从当前点开始找一条允许弧
            if(edge[i].cap&&d[u]==d[edge[i].v]+1)//如果找到跳出循环
                break;

        if(i!=-1)//找到一条允许弧
        {
            cur_edge[u]=i;//从点u出发的允许弧的地址
            pre[edge[i].v]=u;//允许弧上下一个点的前驱为u
            u=edge[i].v;//u变成下一个点继续搜直到搜出一条增广路
        }
        else  //如果没有搜到允许弧
        {
            numd[d[u]]--; //d[u]将被修改所以numd[d[u]]减一
            if(!numd[d[u]]) break;  //如果没有点的d值为d[u]则不可能再搜到增广路结束搜索
            cur_edge[u]=head[u];  //当前点的允许弧为第一条边
            int tmp=n;
            for(int j=head[u]; j!=-1; j=edge[j].next) //搜与u相连的点中d值最小的
                if(edge[j].cap&&tmp>d[edge[j].v])
                    tmp=d[edge[j].v];

            d[u]=tmp+1; //修改d[u]
            numd[d[u]]++;
            if(u!= source)
                u=pre[u];//从u的前驱搜，因为从u没有搜到允许弧
        }
    }
    return max_flow;
}

int main()
{
    //freopen("C:\\Users\\Administrator\\Desktop\\kd.txt","r",stdin);
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d%d",&n,&m);
        int x,y,source,sink;
        int Min=INF, Max=-INF;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&x,&y);
            if(x<=Min)//源点
            {
                source=i;
                Min=x;
            }
            if(x>=Max)//汇点
            {
                sink=i;
                Max=x;
            }
        }
        memset(head,-1,sizeof(head));
        cnt=0;
        int u,v,w;
        for(int i=0; i