地铁网络(分层图-哈密尔顿路径)

来源：互联网发布：淘宝923大促编辑：程序博客网时间：2024/04/26 06:35

最近Csdn的图片似乎有了改革（1.相册见不到了？搬家难度++ 2.水印 3.图片地址长度++ 4.似乎可以用来图床了？TNY）

言归正传：这题是分层图（废话）的加强版——分层图+哈密尔顿路径！！！（几乎没变求D……）

话说我根本不会哈密尔顿啊(求D)，指数级算法a（求D）

接下来讲讲第一次写哈密尔顿（都不屑D我）的Exp：

1.如果你想一个状态由它的真子集递归而来那么（从小到大枚举） Rea:A的真子集<A

2.终点起点一定要拆（以防S,T点拐弯）

3.由2可知初始化和求答案都要考虑2

4.除非k=0，否则S,T点的换乘一定为INF时间

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (50000+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;ll a,b;int n;char s[1000];int x[1000],y[1000];int main(){//freopen("robot.in","r",stdin);//freopen(".out","w",stdout);cin>>a>>b;scanf("%s",s+1);n=strlen(s+1);x[0]=y[0]=0;bool bo=0;For(i,n){if (s[i]=='R') x[i]=x[i-1]+1,y[i]=y[i-1];if (s[i]=='L') x[i]=x[i-1]-1,y[i]=y[i-1];if (s[i]=='U') x[i]=x[i-1],y[i]=y[i-1]+1;if (s[i]=='D') x[i]=x[i-1],y[i]=y[i-1]-1;if (x[i]==a&&y[i]==b) bo=1;}if (bo==1){cout<<"Yes"<<endl;return 0;}Rep(i,n+1){int t1=a-x[i],t2=b-y[i];if (t1!=0&&x[n]==0) continue;if (t2!=0&&y[n]==0) continue;if (x[n]==0||y[n]==0){bool bo1=0,bo2=0;if (x[n]==0) bo1=1;if (y[n]==0) bo2=1;if (bo1&&!bo2) if (t2%y[n]==0&&t2/y[n]>=0) {cout<<"Yes"<<endl;return 0;} else continue; if (!bo1&&bo2) if (t1%x[n]==0&&t1/x[n]>=0) {cout<<"Yes"<<endl;return 0;} else continue;if (!x[i]&&!y[i]) {cout<<"Yes"<<endl;return 0;} else continue;}if (t1%x[n]||t2%y[n]) continue;t1/=x[n],t2/=y[n];if (t1<0||t2<0||t1^t2) continue;cout<<"Yes"<<endl;return 0;}cout<<"No"<<endl;return 0;}