POJ 1606 Jugs(BFS:找最短路径并输出)

POJ 1606 Jugs(BFS:找最短路径并输出)

http://poj.org/problem?id=1606

题意:

        又是给你两个容量为A和B的水杯,要你倒出B杯子有C升水的路径.

分析:

本题之前我就做过一道类似的的:

http://blog.csdn.net/u013480600/article/details/25241777

上面有分析.下面直接给出代码:

AC代码:

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<stack>using namespace std;const int maxr=100+5;const int maxc=100+5;const int maxn=10000+100;//这个规模是测试OJ得出的int A,B,C;int vis[maxr][maxc],dist[maxr][maxc],pre[maxr][maxc];struct Cell{    int a,b;    Cell(int a,int b):a(a),b(b){}};int cnt;struct Node{    int a,b;    int t;    Node(){}    Node(int a,int b,int t):a(a),b(b),t(t){}}nodes[maxn];Cell BFS(){    queue<Cell> Q;    Q.push(Cell(0,0));    memset(vis,0,sizeof(vis));    vis[0][0]=1;    dist[0][0]=0;    cnt=0;    while(!Q.empty())    {        Cell cell=Q.front();Q.pop();        int a=cell.a,b=cell.b;        for(int d=0;d<6;d++)        {            int na,nb;            if(d==0){na=A,nb=b;}            else if(d==1) {na=a,nb=B;}            else if(d==2) {na=0,nb=b;}            else if(d==3) {na=a,nb=0;}            else if(d==4)            {                int all=a+b;                na= all>=B?all-B:0;                nb= all>=B? B:all;            }            else if(d==5)            {                int all=a+b;                na= all>=A? A:all;                nb= all>=A? all-A:0;            }            if(vis[na][nb]==0)            {                vis[na][nb]=1;                dist[na][nb]=dist[a][b]+1;                nodes[cnt++]=Node(a,b,d);                pre[na][nb]=cnt-1;                Q.push(Cell(na,nb));                if(nb==C) return Cell(na,nb);            }        }    }    return Cell(-1,-1);}int main(){    while(scanf("%d%d%d",&A,&B,&C)==3)    {        Cell cell=BFS();        stack<int> S;        int a=cell.a ,b=cell.b;        while(true)        {            int tmp=pre[a][b];            S.push(nodes[tmp].t);            a=nodes[tmp].a,b=nodes[tmp].b;            if(a==0&&b==0) break;        }        while(!S.empty())        {            int t=S.top();S.pop();            if(t==0) printf("fill A\n");            else if(t==1) printf("fill B\n");            else if(t==2) printf("empty A\n");            else if(t==3) printf("empty B\n");            else if(t==4) printf("pour A B\n");            else if(t==5) printf("pour B A\n");        }        printf("success\n");    }    return 0;}