十滴水，半成品，大多数关卡可以过去，不过也有几个过不去，仍在调试中，不断更新

  1 /*
  2 宽搜+记忆化搜索
  3 */
  4 #include <iostream>
  5 #include <cstdio>
  6 #include <string>
  7 #include <queue>
  8 #include <string>
  9 #define  leftedge 1
 10 #define  rightedge    5
 11 #define  topedge  1
 12 #define  downedge 6
 13 using namespace std;
 14 struct status{
 15     int map[7][6];
 16     int step;
 17 };
 18 bool success;
 19 int resx,resy;
 20 int steps[1000];
 21 int limit;
 22 void output(status s){
 23     int i,j;
 24     printf("

---------------
");
 25     for(i=topedge;i<=downedge;++i){
 26         for(j=leftedge;j<=rightedge;++j)
 27             printf("%d ",s.map[i][j]);
 28         printf("
");
 29     }
 30     printf("---------------


");
 31     return;
 32 }
 33 
 34 bool check(status s){
 35     int i,j;
 36     for(i=topedge;i<=downedge;++i)
 37         for(j=leftedge;j<=rightedge;++j){
 38             if(s.map[i][j]!=0) return 0;
 39         }
 40     return 1;
 41 }
 42 struct node{
 43     int x,y;
 44 };
 45 status update(int x,int y,status s){  //已经爆了的气球
 46     queue<node>order;
 47     while(order.size()>0) order.pop();
 48     int i,j,k,ii,jj,tx,ty;
 49     node t,t1,t2;
 50     t.x=x;  t.y=y;
 51     order.push(t);
 52     while(order.size()>0){
 53         t1=order.front();
 54         tx=t1.x;    ty=t1.y;
 55         order.pop();
 56         s.map[tx][ty]=0;
 57         for(i=tx-1;i>=topedge;--i){
 58             if(s.map[i][ty]==0) continue;
 59             else {
 60                 s.map[i][ty]++; t2.x=i; t2.y=ty;
 61                 if(s.map[i][ty]==5) {s.map[i][ty]=0;order.push(t2);}
 62                 break;
 63             }
 64         }
 65         for(i=tx+1;i<=downedge;++i){
 66             if(s.map[i][ty]==0) continue;
 67             else {
 68                 s.map[i][ty]++; t2.x=i; t2.y=ty;
 69                 if(s.map[i][ty]==5) {s.map[i][ty]=0;order.push(t2);}
 70                 break;
 71             }
 72         }
 73         for(i=ty-1;i>=leftedge;--i){
 74             if(s.map[tx][i]==0) continue;
 75             else {
 76                 s.map[tx][i]++; t2.x=tx; t2.y=i;
 77                 if(s.map[tx][i]==5) {s.map[tx][i]=0;order.push(t2);}
 78                 break;
 79             }
 80         }
 81         for(i=ty+1;i<=rightedge;++i){
 82             if(s.map[tx][i]==0) continue;
 83             else {
 84                 s.map[tx][i]++; t2.x=tx; t2.y=i;
 85                 if(s.map[tx][i]==5) {s.map[tx][i]=0;order.push(t2);}
 86                 break;
 87             }
 88         }
 89     }
 90     if(check(s)==1) {resx=x;resy=y; success=1;output(s);}
 91     //output(s);
 92     return s;
 93 }
 94 void change(int step,status s){
 95     int i,j,ii,jj;
 96     status temp;
 97     if(success==1||step>limit||check(s)==1)
 98         return;
 99     for(i=topedge;i<=downedge;++i)
100         for(j=leftedge;j<=rightedge;++j){
101             temp=s;
102             if(temp.map[i][j]==0) continue;
103             temp.map[i][j]++;
104 
105             temp.step=i*10+j;
106             steps[step*100+temp.step]=s.step;
107 
108 
109 
110             if(temp.map[i][j]==5) temp=update(i,j,temp);
111             if(success==1){
112                 return;
113             }
114             change(step+1,temp);
115             if(success==1){
116                 printf("ok
");
117 
118 
119 
120 
121                 return ;
122             }
123 
124 
125 
126         }
127     return;
128 }
129 int main(){
130     int i,j,tx,ty;
131     status s;
132     int res[10];
133     while(true){
134         success=0;
135         printf("请输入限制步数
");
136         scanf("%d",&limit);
137         for(i=topedge;i<=downedge;++i)
138             for(j=leftedge;j<=rightedge;++j)
139                 scanf("%d",&s.map[i][j]);
140         s.step=0;
141 
142         change(1,s);
143         printf("success=%d
",success);
144 
145 
146         res[limit]=limit*100+resx*10+resy;
147         tx=resx;    ty=resy;
148 
149         for(i=limit-1;i>=1;--i){
150             res[i]= steps[ (i+1)*100+ tx*10+ty  ];
151             tx=res[i]/10;   ty=res[i]%10;
152         }
153         for(i=1;i<=limit;i++)
154         {
155             res[i]=res[i]%100;
156             printf("%d %d
",res[i]/10,res[i]%10);
157         }
158     }
159     return 0;
160 }