poj 2331

（ＩＤＡ＊：　即为A*算法加上迭代加深搜索，本题为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

我的IDA*算法的第一题^-^...

它与bfs不同的地方是前面多用的一个“限制”，在本题中，既是先不管水管的个数，假设所有水管无限多。分别看横纵坐标从终点倒着走到（一维，因为横纵坐标分别考虑）每个点所需的步数。然后利用这个进行迭代加深搜索(枚举答案)与大幅度剪枝.

具体细节见代码：

<span style="color:#000000;">#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>using namespace std;#define MAXN (1000+5)struct node{int l, c;};int n, ans;int sx, sy, tx, ty;node pipe[11];int hx[MAXN], hy[MAXN];void bfs(int *h, int x){queue<int> q;h[x] = 0;q.push(x);while(!q.empty()){int now = q.front(); q.pop();for(int i = 1; i <= n; i++){int len = pipe[i].l;if(now-len >= 1 && h[now-len] == -1){h[now-len] = h[now]+1;q.push(now-len);}if(now+len <= 1000 && h[now+len] == -1){h[now+len] = h[now]+1;q.push(now+len);}}}}bool check(node *a, int x, int dep, int k){int hv = k? hy[x]: hx[x];if(hv+dep > ans || hv == -1) return false;if(!hv){if(!k) return check(a, sy, dep, 1);else return true;}node tmp[11];for(int i = 1; i <= n; i++) tmp[i] = a[i];for(int i = 1; i <= n; i++)    if(tmp[i].c){tmp[i].c--;if((x-tmp[i].l)>=1 && check(tmp, x-tmp[i].l, dep+1, k)) return true;if((x+tmp[i].l)>=1 && check(tmp, x+tmp[i].l, dep+1, k)) return true;tmp[i].c++;}return false;}int main(){scanf("%d%d%d%d", &sx, &sy, &tx, &ty);scanf("%d", &n);for(int i = 1; i <= n; i++) scanf("%d", &pipe[i].l);int num = 0;for(int i = 1; i <= n; i++){scanf("%d", &pipe[i].c);num += pipe[i].c;}memset(hx, -1, sizeof(hx));memset(hy, -1, sizeof(hy));bfs(hx, tx);bfs(hy, ty);for(ans = 1; ans <= num; ans++)  if(check(pipe, sx, 0, 0)) break;if(ans <= num) printf("%d\n", ans);else printf("-1\n");return 0;}</span>