poj2331 Water pipe IDA*

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

题目大意
在平面上有一个起点和一个终点，你需要用一些给定的管子将这两个点连起来，求最少需要多少管子。

输入
首先5个整数sx,sy,tx,ty,n。sx，sy代表起点坐标；tx，ty代表终点坐标；
n代表管子数量。接下来n个整数分别代表每个管子的长度，在接着n个整数分别代表每个管子的数量。

输出
一个整数代表最少的管子数。

总结
昨天学了A*和IDA*启发式搜索，一直在改这道题，后来发现就是dfs没有回溯，弄了我将近两个小时。我觉得有些无语。

先简要谈一下对A*以及IDA*的看法。
A*以及IDA*都是启发式搜索，他们都是在一般搜索的基础上增加了启发式函数（估价函数），对于一个启发式搜索，估价函数是非常重要的，估价函数通俗来讲就是一个估算到目的地要花费多少，有了一个好的估价函数，就少了很多不必要的搜索。相反，一个错误估价函数，会将你的搜索引入歧途。既然估价函数对我们这么重要，那我们应该怎么求呢？其实在不同的题目之中有不同的估价函数，比较常见的就是曼哈顿函数，通过计算两点之间绝对值之差来算。

那我们又要问A*与IDA*之间有什么区别呢？
其实很简单。A*是由bfs加估价函数构成的，而IDA*是由dfs加估价函数构成的，由于dfs不能控制深度，我们需要通过枚举深度来算，一旦出现解，必定是最优解。

回到这一道题，这一道题我们的估价函数是通过bfs不计线段数量算出来的，而实际情况是大于等于这种情况的。

WA的原因
一直b[i].num–之后忘记++了——

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<queue>using namespace std;const int maxn=1001,inf=1e9;int sx,sy,tx,ty,sum,ans,h1[maxn],h2[maxn],n;struct node{    int len,num;};bool dfs1(node *a,int x,int z){    int i,j;    if(h2[x]==-1 || h2[x]+z>ans) return false;    if(h2[x]==0){        //printf("->h2--%d___%d\n",x,z);        return true;    }    node b[11];    for(i=1;i<=n;i++)        b[i]=a[i];    for(i=1;i<=n;i++)        if(b[i].num>0){            b[i].num--;            if(x-b[i].len>=1)                if(dfs1(b,x-b[i].len,z+1)){                    //printf("->h2--%d\n",x);                    return true;                  }            if(x+b[i].len<=1000)                if(dfs1(b,x+b[i].len,z+1)){                    //printf("->h2--%d\n",x);                     return true;                }            b[i].num++;        }    return false;}bool dfs(node *a,int x,int z){    int i,j;    if(h1[x]==-1 || h1[x]+z>ans) return false;    if(h1[x]==0){        //printf("%d__%d\n",x,z);        //for(i=1;i<=n;i++) printf("%d-----",a[i].num);        return dfs1(a,sy,z);    }    node b[11];    for(i=1;i<=n;i++)        b[i]=a[i];    for(i=1;i<=n;i++)        if(b[i].num>0){            b[i].num--;            if(x-b[i].len>=1)                if(dfs(b,x-b[i].len,z+1)){                    //printf("->h1--%d----id:%d\n",x,i);                    return true;                  }            if(x+b[i].len<=1000)                if(dfs(b,x+b[i].len,z+1)){                    //printf("->h1--%d----id:%d=====x+b[i]=%d\n",x,i,x+b[i].len);                     return true;                }            b[i].num++;        }    return false;}node a[11];void bfs(int *h,int st){    int i,j;    queue <int> q;    h[st]=0;    q.push(st);    while(!q.empty()){        int x;        x=q.front();        q.pop();        for(i=1;i<=n;i++){            if(x-a[i].len>=1)                if(h[x-a[i].len]==-1){                    h[x-a[i].len]=h[x]+1;                    q.push(x-a[i].len);                }            if(x+a[i].len<=1000)                if(h[x+a[i].len]==-1){                    h[x+a[i].len]=h[x]+1;                    q.push(x+a[i].len);                }        }    }}void id_astar(){    int i;    if(sx==tx && sy==ty){        puts("0");        return;    }    memset(h1,-1,sizeof(h1));    memset(h2,-1,sizeof(h2));    bfs(h1,tx);    bfs(h2,ty);    /*for(i=1;i<=1000;i++)        if(h1[i]!=-1 || h2[i]!=-1)            printf("h1[%d]=%d    h2[%d]=%d\n",i,h1[i],i,h2[i]);*/    for(ans=1;ans<=sum;ans++){        //if(ans==4)            //system("pause");        if(dfs(a,sx,0))            break;    }    if(ans<=sum)        printf("%d\n",ans);    else puts("-1");    return;}   int main(){    int i,j,k;    int x1,y1,x2,y2;    //freopen("test.in","r",stdin);    //freopen("test.out","w",stdout);    scanf("%d%d%d%d%d",&sx,&sy,&tx,&ty,&n);    for(i=1;i<=n;i++)        scanf("%d",&a[i].len);    for(i=1;i<=n;i++){        scanf("%d",&a[i].num);        sum+=a[i].num;    }    id_astar();    return 0;}