本来用string来记录状态，tle，于是改为用类似于并查集的方法记录。只记录 此次动作 和 上一个动作的位置。

中间又有一些低级失误，wa了n次。。orz

/* ID: immozer1 PROG: **** LANG: C++ */ #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <fstream> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <cstring> using namespace std; struct Status { int l,r; }; Status st[20000]; //记录此次的动作 int rec[20000]; bool v[20000]; //用了类似并查集的方法记录上一个动作的位置 int father[20000]; int bfs(int a,int b,int c) { int top,end,tl,tr; top = end = 0; st[end].l = 0; father[end] = end; st[end++].r = 0; v[0] = true; while(top < end) { if(st[top].l == c || st[top].r == c) return top; tl = st[top].l; tr = st[top].r; //分别对应fill(1),fill(2),drop(1),…,pour(2,1) if(tl != a && !v[150*a+tr]) { v[150*a+tr] = true; st[end].l = a; st[end].r = tr; father[end] = top; rec[end++] = 1; } if(tr != b && !v[150*tl+b]) { v[150*tl+b] = true; st[end].l = tl; st[end].r = b; father[end] = top; rec[end++] = 2; } if(tl != 0 && !v[tr]) { v[tr] = true; st[end].l = 0; st[end].r = tr; father[end] = top; rec[end++] = 3; } if(tr != 0 && !v[150*tl]) { v[150*tl] = true; st[end].l = tl; st[end].r = 0; father[end] = top; rec[end++] = 4; } int tmp; if(tl != a && tr != 0) { tmp = min(a-tl,tr); if(!v[(tl+tmp)*150+tr-tmp]) { v[(tl+tmp)*150+tr-tmp] = true; st[end].l = tl + tmp; st[end].r = tr – tmp; father[end] = top; rec[end++] = 5; } } if(tl != 0 && tr != b) { tmp = min(tl,b-tr); if(!v[(tl-tmp)*150+tr+tmp]) { v[(tl-tmp)*150+tr+tmp] = true; st[end].l = tl – tmp; st[end].r = tr + tmp; father[end] = top; rec[end++] = 6; } } top++; } return 0; } int main() { int a,b,c; cin >> a >> b >> c; int sp[10000],end,now; now = bfs(a,b,c); end = 0; //将并查集的元素取出,记录 while(now != father[now]) { sp[end++] = now; now = father[now]; } if(end == 0) { cout << “impossible” << endl; return 0; } string y; cout << end << endl; for(int i = end-1; i >= 0; i–) { switch(rec[sp[i]]) { case 1: y = “FILL(1)”; break; case 2: y = “FILL(2)”; break; case 3: y = “DROP(1)”; break; case 4: y = “DROP(2)”; break; case 5: y = “POUR(2,1)”; break; default: y = “POUR(1,2)”; } cout << y << endl; } return 0; }