H – PotsTime Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 3414DescriptionYou are given two pots, having the volume of A and B liters respectively. The following operations can be performed:FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;DROP(i) empty the pot i to the drain;POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.InputOn the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).OutputThe first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.Sample Input3 5 4Sample Output6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)
POUR(2,1)
//刘汝佳的小白书里，只是提到过一个概念，叫做隐式图搜索，个人觉得，就是搜索的，不是常见的矩阵或者邻接表存储的图，而是比较抽象的图，以本题为例，状态可以理解为图的节点，由这个状态能到另一个状态就是表明这两个状态之间是有边的 ，那么就可以以图的搜索思路来写了，本题的状态是两个杯子的水量，，比如初始没有水，那就是（0,0），但是有三种操作，所以由（0,0）可以到（A，0）,(0,B)…等
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<map>#include<stack>using namespace std;const int maxn=50005;const int inf=200000;#define lson rt<<1,l,m#define rson rt<<1|1,m+1,rtemplate<class T>inline T read(T&x){char c;while((c=getchar())<=32);bool ok=false;if(c=='-')ok=true,c=getchar();for(x=0; c>32; c=getchar())x=x*10+c-'0';if(ok)x=-x;return x;}template<class T> inline void read_(T&x,T&y){read(x);read(y);}template<class T> inline void write(T x){if(x<0)putchar('-'),x=-x;if(x<10)putchar(x+'0');else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){write(x);putchar('\n');}// ——-IO template——int A,B,C;struct node{int a,b;int id;node(int x,int y,int d){id=d;a=x;b=y;}};bool vis[105][105];int num=0;int d[maxn];int p[maxn];bool ok;map<int,string> mp;map<int,string>::iterator it;void bfs(node s){queue<node> q;q.push(s);memset(vis,false,sizeof(vis));memset(p,-1,sizeof(p));mp.clear();d[0]=0;p[0]=-1;vis[s.a][s.b]=true;num=1;while(!q.empty()){node s=q.front();q.pop();if(s.a==C||s.b==C){printf("%d\n",d[s.id]);it=mp.begin();stack<string> ss;for(int i=s.id;i!=-1;i=p[i]){// printf("(%d)\n",i);it=mp.begin();while(it->first!=i&&it!=mp.end()){it++;}ss.push(it->second);}ss.pop();while(!ss.empty()){printf("%s\n",ss.top().c_str());ss.pop();}ok=1;return;}//bfs按层次搜索能到达的六条边 if(s.a<A&&!vis[A][s.b]){vis[A][s.b]=true;q.push(node(A,s.b,num));d[num]=d[s.id]+1;mp[num]="FILL(1)";p[num++]=s.id;}if(s.b<B&&!vis[s.a][B]){vis[s.a][B]=true;q.push(node(s.a,B,num));d[num]=d[s.id]+1;mp[num]="FILL(2)";p[num++]=s.id;}if(s.a>0&&!vis[0][s.b]){vis[0][s.b]=true;q.push(node(0,s.b,num));d[num]=d[s.id]+1;mp[num]="DROP(1)";p[num++]=s.id;}if(s.b>0&&!vis[s.a][0]){vis[s.a][0]=true;q.push(node(s.a,0,num));d[num]=d[s.id]+1;mp[num]="DROP(2)";p[num++]=s.id;}if(s.a>0&&s.b<B){int t=min(s.a,B-s.b);if(!vis[s.a-t][s.b+t]){vis[s.a-t][s.b+t]=true;q.push(node(s.a-t,s.b+t,num));d[num]=d[s.id]+1;mp[num]="POUR(1,2)";p[num++]=s.id;}}if(s.a<A&&s.b>0){int t=min(s.b,A-s.a);if(!vis[s.a+t][s.b-t]){vis[s.a+t][s.b-t]=true;q.push(node(s.a+t,s.b-t,num));d[num]=d[s.id]+1;mp[num]="POUR(2,1)";p[num++]=s.id;}}}}int main(){int n,m,i,j,k,t; // freopen("in.txt","r",stdin);read(A);read(B);read(C);ok=0;bfs(node(0,0,0));if(!ok)printf("impossible\n");return 0;}
带上心灵去旅行，以平和的心态看待一切，
