uva1025(基础dp)

思路：先要与处理出来每辆列车经过每个站点的时间方向。再然后就是一通模拟了。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
inline int Readint(){
	char c = getchar();
	while(!isdigit(c)) c = getchar();
	int x = 0;
	while(isdigit(c)){
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x;
}
const int maxn = 1e3 + 10;
int mx[maxn],my[maxn],dp[maxn][maxn],t[maxn];
bool is[maxn][maxn][2];
int m1,m2;
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int n,icase = 0;
	while(scanf("%d",&n) != EOF && n){
		// cout << "n = " << n << endl;
		memset(is, false,sizeof is);
		int tt;

		scanf("%d",&tt);
		for (int i = 1;i < n;i++)
			cin >> t[i];

		scanf("%d",&m1);
		for (int i = 1;i <= m1;++i)
			scanf("%d",&mx[i]);

		scanf("%d",&m2);
		for (int j = 1;j <= m2;++j)
			scanf("%d",&my[j]);

		for (int j = 1;j <= m1;++j){
			int sum = 0;
			is[mx[j]][1][0] = true;
			for (int i = 1;i < n;i++){
				sum += t[i];
				is[sum + mx[j]][i + 1][0] = true;
			}
		}

		for (int j = 1;j <= m2;++j){
			int sum = 0;
			is[my[j]][n][1] = true;
			for (int i = n - 1;i > 0;--i){
				sum += t[i];
				is[my[j] + sum][i][1] = true;
			}
		}
		memset(dp, 0,sizeof dp);
		for (int i = 1;i <= n - 1;i++) dp[tt][i] = inf;
		dp[tt][0] = 0;
		for (int i = tt - 1;i >= 0;--i){
			for (int j = 1;j <= n;j++){
				dp[i][j] = dp[i + 1][j] + 1;
				if (j < n && is[i][j][0] && i + t[j] <= tt){
					dp[i][j] = min(dp[i][j],dp[i + t[j]][j + 1]);
				}
				if (j > 1 && is[i][j][1] && i + t[j - 1] <= tt){
					dp[i][j] = min(dp[i][j],dp[i + t[j - 1]][j - 1]);
				}
			}
		}
		printf("Case Number %d: ",++icase);
		if (dp[0][1] >= INF) puts("impossible");
		else printf("%d\n",dp[0][1]);
	}
	return 0;
}