題目大意：给出2个水杯和目标的容量，输出经过fill drop pour三种操作的最少步数。
这题思路还是比较清晰的，可是就是不知为什么mle了，找不出来问题，先码着放这里，有空回来看。
mle:
#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<map>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
int a,b,c,cnt[105][105],faedge[105][105],mins,ans;
bool vis[105][105];struct tie{int ra,rb;tie(int aa,int bb):ra(aa),rb(bb){}tie(){}
};
map<int,map<int,tie> > fanode;void print(tie t){map<int,string> m;stack<tie> s;m[0]="FILL(1)";m[1]="FILL(2)";m[2]="DROP(1)";m[3]="DROP(2)";m[4]="POUR(1,2)";m[5]="POUR(2,1)";for(;;){if(t.ra==a&&t.rb==b) break;s.push(t);t=fanode[t.ra][t.rb];}while(!s.empty()){tie temp=s.top();s.pop();cout<<m[faedge[temp.ra][temp.rb]]<<endl;}
}bool bfs(tie t){queue<tie> q;q.push(t);vis[t.ra][t.rb]=true;cnt[t.ra][t.rb]=0;while(!q.empty()){tie temp=q.front();q.pop();if(a-temp.ra==c||b-temp.rb==c){cout<<cnt[temp.ra][temp.rb]<<endl;print(temp);return true;}for(int i=0;i<6;i++){if(i==0){///fill(1)if(temp.ra){int rra=0;int rrb=temp.rb;if(!vis[rra][rrb]){vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}else if(i==1){///fill(2)if(temp.rb){int rrb=0;int rra=temp.ra;if(!vis[rra][rrb]){vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}else if(i==2){///drop(1)if(temp.ra!=a){int rrb=temp.rb;int rra=a;if(!vis[rra][rrb]){vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}else if(i==3){///drop(2)if(temp.rb!=b){int rrb=b;int rra=temp.ra;if(!vis[rra][rrb]){vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}else if(i==4){///pour(1,2)int rra,rrb;if(temp.ra!=a&&temp.rb){///a不为空且b有空位int aw=a-temp.ra;///a瓶中的酒if(aw>temp.rb){rrb=0;rra=temp.ra+temp.rb;}else{rra=a;rrb=temp.rb-aw;}if(!vis[rra][rrb]){vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}else if(i==5){///pour(2,1)int rra,rrb;if(temp.rb!=b&&temp.ra){///b不为空且a有空位int bw=b-temp.rb;///b瓶中的酒if(bw>temp.ra){rra=0;rrb=temp.rb+temp.ra;}else{rrb=b;rra=temp.ra-bw;}vis[rra][rrb]=true;cnt[rra][rrb]=cnt[temp.ra][temp.rb]+1;faedge[rra][rrb]=i;fanode[rra][rrb].ra=temp.ra;fanode[rra][rrb].rb=temp.rb;q.push(tie(rra,rrb));}}}}return false;
}int main(){while(scanf("%d%d%d",&a,&b,&c)!=EOF){memset(vis,0,sizeof(vis));memset(cnt,-1,sizeof(cnt));memset(faedge,0,sizeof(faedge));tie t;t.rb=b;t.ra=a;if(bfs(t));elsecout<<"impossible"<<endl;}return 0;
}
ac(爬的别人的，感觉差不多啊）:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 105;
const double eps = 1e-7;
bool vis[maxn][maxn];
const int dir[4][2]= {1, 0, 0, 1, -1, 0, 0, -1};
char map[maxn][maxn];int a, b, k;
struct node
{int vola, volb, step;char str[maxn][maxn];
};bool bfs()
{memset(vis, false, sizeof(vis));queue<node> que;node p, q;p.vola = 0, p.volb = 0, p.step = 0;que.push(p);vis[0][0] = 1;///vis[p.vola][p.volb] = true;while(!que.empty()){p = que.front();que.pop();if(p.vola==k || p.volb == k){cout<<p.step<<endl;for(int i=1; i<=p.step; i++)///cout<<p.str[i]<<endl;printf("%s\n",p.str[i]);return true;}///倒满 aif(p.vola == 0){q = p;q.vola = a;q.step++;strcpy(q.str[q.step], "FILL(1)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}}///把 a 中的水倒出来else if(p.vola <= a){q = p;q.vola = 0;q.step++;strcpy(q.str[q.step], "DROP(1)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}///a -> bif(p.volb < b){q = p;if(q.volb + q.vola <= b){q.volb += q.vola;q.vola = 0;}else{q.vola = (q.vola+q.volb)-b;q.volb = b;}q.step++;strcpy(q.str[q.step], "POUR(1,2)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}}}///把 b 倒满if(p.volb == 0){q = p;q.volb = b;q.step++;strcpy(q.str[q.step], "FILL(2)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}}///把 b 中的水倒出来else if(p.volb <= b){q = p;q.volb = 0;q.step++;strcpy(q.str[q.step],"DROP(2)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}if(p.vola < a){q = p;if(q.vola + q.volb <= a){q.vola += q.volb;q.volb = 0;}else{q.volb = (q.vola+q.volb)-a;q.vola = a;}q.step++;strcpy(q.str[q.step], "POUR(2,1)");if(!vis[q.vola][q.volb]){vis[q.vola][q.volb] = true;que.push(q);}}}}return false;
}
int main()
{while(cin>>a>>b>>k){bool ok = bfs();if(!ok)puts("impossible");}return 0;
}