Water pipe
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 2265 Accepted: 602

Description
The Eastowner city is perpetually haunted with water supply shortages, so in order to remedy this problem a new water-pipe has been built. Builders started the pipe from both ends simultaneously, and after some hard work both halves were connected. Well, almost. First half of pipe ended at a point (x1, y1), and the second half – at (x2, y2). Unfortunately only few pipe segments of different length were left. Moreover, due to the peculiarities of local technology the pipes can only be put in either north-south or east-west direction, and be connected to form a straight line or 90 degree turn. You program must, given L1, L2, … Lk – lengths of pipe segments available and C1, C2, … Ck – number of segments of each length, construct a water pipe connecting given points, or declare that it is impossible. Program must output the minimum required number of segments.

Constraints
1 <= k <= 4, 1 <= xi, yi, Li <= 1000, 1 <= Ci <= 10

Input
Input contains integers x1 y1 x2 y2 k followed by 2k integers L1 L2 … Lk C1 C2 … Ck

Output
Output must contain a single integer – the number of required segments, or ?1 if the connection is impossible.

Sample Input

20 10 60 50 2 70 30 2 2

Sample Output

4

Source
Northeastern Europe 2003, Far-Eastern Subregion

题目链接：

id=2331">http://poj.org/problem?id=2331
题意：
在二维网格上给你起点，终点，与（n《=10）的管子（长度与数量）用最少的管子数完毕路径；

思路：由于管子不能切断。所以“盲目”dfs 长路漫漫。。。
仅仅好迭代加深！

——————–分开计算x。y轴————————–
1.预处理*h数组。计算i状态到终点（单维）的最最短路（估价系统）
for(ans=1;ans<=tot;ans++){if(dfs(a,sx,0,0)) break;}

2.“盲目”dfs（+剪枝）到终点（单维）
剪*
if(hv==-1||hv+dep>ans) return 0;

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct node{
    int l,num;
}a[11];
int n,h1[1010],h2[1010],ans;
int tx,ty,sx,sy;
int tot;

void cal(int *h,int pos)
{
    queue<int> q;
    h[pos]=0;
    q.push(pos);
    while(!q.empty())
    {
        pos=q.front();
        q.pop();
        for(int i=1;i<=n;i++)
        {
            int nex=pos-a[i].l;
            if(nex>0&&h[nex]==-1)
            {
                h[nex]=h[pos]+1;
                q.push(nex);
            }
            nex+=2*a[i].l;
            if(nex<=1000&&h[nex]==-1)
            {
                h[nex]=h[pos]+1;
                q.push(nex);        
            }
        }
    }
}
bool dfs(node *a,int x,int dep,int id)
{
    int hv;
    if(id==0) hv=h1[x];else hv=h2[x];
    if(hv==-1||hv+dep>ans) return 0;
    if(hv==0)
    {
        if(id==0) return dfs(a,sy,dep,1);
        else return 1;
    }

    node tmp[10];
    for(int i=1;i<=n;i++) tmp[i]=a[i];
    for(int i=1;i<=n;i++)
    if(tmp[i].num)
    {
        tmp[i].num--;

        int now=x-tmp[i].l;
        if(now>0) if(dfs(tmp,now,dep+1,id)) return 1;
        now+=2*tmp[i].l;
        if(now<=1000) if(dfs(tmp,now,dep+1,id)) return 1;
        tmp[i].num++;
    }
    return 0;

}



void id_a_star()
{
    memset(h1,-1,sizeof(h1));
    memset(h2,-1,sizeof(h2));
    cal(h1,tx);
    cal(h2,ty);

    for(ans=1;ans<=tot;ans++)
    {
        if(dfs(a,sx,0,0)) break;
    }
    if(ans<=tot)
    printf("%d\n",ans);
    else printf("-1\n");

}
int main()
{
    scanf("%d%d%d%d%d",&sx,&sy,&tx,&ty,&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i].l);
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&a[i].num);
        tot+=a[i].num;
    }

    if(sx==tx&&sy==ty) printf("0\n");
    else id_a_star();


}