H - Pots POJ - 3414

题意： (两个锅互相倒水)希望能通过题目中的几种操作使一个锅中的水量为目标水量
bfs 搜索每一步的每种操作，直到所有操作用完，则impossible,否则输出答案
为了方便操作，用两个map，一个存六种操作对应的结果输出，一个当搜索时的visit函数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
using namespace std;
const int maxn = 1e5;
int n, m, goal;
struct node {int a1, a2; // 两锅中的剩余水量 int step; // 操作次数 int op[maxn];
};map<int, string> opera;
map<pair<int, int>,int> mp;node bfs() {node s;s.a1 = 0; s.a2 = 0; s.step = 0;mp[make_pair(0,0)] = 1;queue<node> Q;Q.push(s);while(!Q.empty()) {node now = Q.front(); Q.pop();if(now.a1 == goal || now.a2 == goal) return now;//      cout << "*" << now.a1 << " " << now.a2 << " " << now.op[now.step] << " " << now.step << endl; for(int i = 1; i <= 6; i++) {node tmp= now; tmp.step = now.step + 1;if(i == 1 && now.a1 < n) {tmp.a1 = n; tmp.op[tmp.step] = 1;if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}else if(i == 2 && now.a2 < m) {tmp.a2 = m; tmp.op[tmp.step] = 2; if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}else if(i == 3 && now.a1 > 0) {tmp.a1 = 0; tmp.op[tmp.step] = 3; if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}else if(i == 4 && tmp.a2 > 0) {tmp.a2 = 0; tmp.op[tmp.step] = 4; if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}else if(i == 5 && tmp.a1 > 0 && tmp.a2 < m) {int temp2 = m - tmp.a2; tmp.op[tmp.step] = 5; if(temp2 > tmp.a1) tmp.a2 += tmp.a1, tmp.a1 = 0;else tmp.a1 -= temp2, tmp.a2 = m;if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}else if(i == 6 && tmp.a1 < n && tmp.a2 > 0) {int temp1 = n - tmp.a1; tmp.op[tmp.step] = 6;if(temp1 > tmp.a2) tmp.a1 += tmp.a2, tmp.a2 = 0;else tmp.a2 -= temp1, tmp.a1 = n;if(!mp[make_pair(tmp.a1, tmp.a2)]) mp[make_pair(tmp.a1, tmp.a2)] = 1, Q.push(tmp);}}}return s;
}int main() {//  freopen("test.in", "r", stdin);ios::sync_with_stdio(false);opera[1] = "FILL(1)";opera[2] = "FILL(2)";opera[3] = "DROP(1)";opera[4] = "DROP(2)";opera[5] = "POUR(1,2)";opera[6] = "POUR(2,1)";while(~scanf("%d%d%d", &n, &m, &goal)) {node ans;ans = bfs();if(ans.step == 0) cout << "impossible" << endl;else {cout << ans.step << endl;for(int i = 1; i <= ans.step; i++) {cout << opera[ans.op[i]] << endl;}}}return 0;
}

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