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(); }

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