poj3414 pots(经典搜索题目， 题目本身不难，处理有点繁琐）

#include <iostream>#include <cstdio>#include <cstring>#include  <queue>using namespace std;int a, b, c;int vist[105][105];struct node {    int x, y;    int id;    int op;    int num;    int pre;}s[100005];int cnt;node cur , next;node judge(node cur, int k) {  //6种状态的处理代码；    if(k == 1)        cur.x = 0;    else if(k == 2)       cur.y = 0;    else if(k == 3)        cur.x = a;    else if(k == 4) {        cur.y = b;    }else if(k == 5) {        if(b - cur.y < cur.x) {            cur.x = cur.x - (b - cur.y);            cur.y = b;        }else {            cur.y += cur.x;            cur.x = 0;        }    }else {        if(a - cur.x < cur.y) {            cur.y = cur.y - (a - cur.x);            cur.x = a;        }else {            cur.x += cur.y;            cur.y = 0;        }    }    return cur;}int dfs(int ta, int tb) {    queue<node>que;    s[0].id = 0;    s[0].x = ta;    s[0].y = tb;    s[0].num = 0;    s[0].op = 0;    s[0].pre = 0;    cnt++;    vist[ta][tb] = 1;    que.push(s[0]);    while(!que.empty()) {        next = que.front();        que.pop();       // printf("%d  %d   %d  %d\n", next.x, next.y, next.pre, next.id);        if(next.x == c || next.y == c) {            return next.id;        }        for(int i = 1; i <= 6; i++) {          //  printf("%d  %d, %d\n", cur.x, cur.y, i);            cur = judge(next, i);            //  printf("h %d  %d, %d\n", cur.x, cur.y, i);            if(!vist[cur.x][cur.y]) {             //    printf("hh %d  %d\n", cur.x, cur.y);                s[cnt].pre = next.id;                s[cnt].id = cnt;                s[cnt].x = cur.x;                s[cnt].y = cur.y;                s[cnt].num = cur.num + 1;                s[cnt++].op = i;               que.push(s[cnt-1]);               vist[cur.x][cur.y] = 1;            }        }    }    return -1;}void put(int i) {  //处理的6个状态对应的操作输出。        if(i == 1) {            printf("DROP(1)\n");        }else if(i == 2) {            printf("DROP(2)\n");        }else if(i == 3) {            printf("FILL(1)\n");        }else if(i == 4) {            printf("FILL(2)\n");        }else if(i == 5) {            printf("POUR(1,2)\n");        }else {            printf("POUR(2,1)\n");        }    }void output(int k) {  //递归输出结果， 即反推回去；    if(s[k].pre == 0) {        put(s[k].op);    }else{        output(s[k].pre);        put(s[k].op);    }}int main(){   while(scanf("%d%d%d", &a, &b, &c) != EOF) {        memset(vist, 0, sizeof(vist));        cnt = 0;        int ans = dfs(0, 0);        if(ans == -1)           printf("impossible\n");        else {                printf("%d\n", s[ans].num);            output(ans);        }   }    return 0;}