看到这个题,接着下手写了个bfs,华丽丽的AC了,真的很难得,不过看题解他们都是dfs?晕
顺便庆祝usaco又过了一个小小节.........
code
/* ID: yueqiq PROG: milk3 LANG: C++ */ #include <set> #include <map> #include <ctime> #include <queue> #include <cmath> #include <stack> #include <limits> #include <vector> #include <bitset> #include <string> #include <cstdio> #include <cstring> #include <fstream> #include <string.h> #include <iostream> #include <algorithm> #define Si set<int> #define LL long long #define pb push_back #define PS printf(" ") #define Vi vector<int> #define LN printf("\n") #define lson l,m,rt << 1 #define rson m+1,r,rt<<1|1 #define SD(a) scanf("%d",&a) #define PD(a) printf("%d",a) #define SET(a,b) memset(a,b,sizeof(a)) #define FF(i,a) for(int i(0);i<(a);i++) #define FD(i,a) for(int i(a);i>=(1);i--) #define FOR(i,a,b) for(int i(a);i<=(b);i++) #define FOD(i,a,b) for(int i(a);i>=(b);i--) #define readf freopen("milk3.in","r",stdin) #define writef freopen("milk3.out","w",stdout) const int maxn = 200000; const int INF = 1111; const int dx[]={0,1,0,-1}; const int dy[]={1,0,-1,0}; const double pi = acos(-1.0); using namespace std; int A,B,C; bool vis[3000][3000]; struct node{ int a,b,c; }q[maxn]; int res[maxn],cnt; node op1(node x){ //将c桶牛奶倒入a桶 node tmp; int t=min(x.c,A-x.a); tmp.a=t+x.a; tmp.b=x.b; tmp.c=x.c-t; return tmp; } node op2(node x){//将c桶牛奶导入b桶 node tmp; int t=min(x.c,B-x.b); tmp.b=t+x.b; tmp.a=x.a; tmp.c=x.c-t; return tmp; } node op3(node x){//将b桶牛奶倒入a桶 node tmp; int t=min(x.b,A-x.a); tmp.a=t+x.a; tmp.c=x.c; tmp.b=x.b-t; return tmp; } node op4(node x){//将b桶牛奶倒入c桶 node tmp; int t=min(x.b,C-x.c); tmp.a=x.a; tmp.c=x.c+t; tmp.b=x.b-t; return tmp; } node op5(node x){//将a桶牛奶倒入c桶 node tmp; int t=min(x.a,C-x.c); tmp.a=x.a-t; tmp.b=x.b; tmp.c=x.c+t; return tmp; } node op6(node x){//将a桶牛奶倒入b桶 node tmp; int t=min(x.a,B-x.b); tmp.a=x.a-t; tmp.b=x.b+t; tmp.c=x.c; return tmp; } void bfs(){ int front,rear; front=rear=0; q[rear].a=0; q[rear].b=0; q[rear++].c=C; vis[0][0]=true; while(front<rear){ node tmp=q[front++]; if(tmp.a==0){ res[cnt++]=tmp.c; } node tmp2=op1(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } tmp2=op2(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } tmp2=op3(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } tmp2=op4(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } tmp2=op5(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } tmp2=op6(tmp); if(!vis[tmp2.a][tmp2.b]){ q[rear++]=tmp2; vis[tmp2.a][tmp2.b]=true; } } } int main() { readf; writef; scanf("%d%d%d",&A,&B,&C); bfs(); sort(res,res+cnt); FOR(i,0,cnt-2){ PD(res[i]);PS; } PD(res[cnt-1]);LN; return 0; }