UVA 1025

紫书有详细的思路，题目挺水的，不再赘述

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int INF = 0x7f7f7f7f;
int n, T;
int t[55];
bool has_train[300][55][2];
int m, x;
int dp[300][55];
int kase=0;

int main(){
    while(scanf("%d", &n), n){
        scanf("%d", &T);
        for(int i = 1; i < n; ++i)
            scanf("%d", t+i);
        memset(has_train, false, sizeof(has_train));

        scanf("%d", &m);
        while(m--){
            scanf("%d", &x);
            has_train[x][1][0]=true;
            for(int i=1; i<n; ++i){
                x+=t[i];
                if(x<300) has_train[x][i+1][0]=true;
            }
        }

        scanf("%d", &m);
        while(m--){
            scanf("%d", &x);
            has_train[x][n][1]=true;
            for(int i=n-1; i>0; --i){
                x+=t[i];
                if(x<300) has_train[x][i][1]=true;
            }
        }

        for(int i =1; i <= n-1; ++i) dp[T][i] = INF;
        dp[T][n] = 0;

        for(int i  = T-1; i >= 0 ; --i)
        for(int j =1; j <= n; ++j){
            dp[i][j] = dp[i+1][j] + 1;
            if(j < n && has_train[i][j][0] && i +t[j] <= T)
                dp[i][j] = min(dp[i][j], dp[i+t[j]][j+1]);
            if(j > 1 && has_train[i][j][1] && i+t[j-1] <= T )
                dp[i][j] = min(dp[i][j], dp[i+t[j-1]][j-1]);
        }

        printf("Case Number %d: ", ++kase);

        if(dp[0][1] >= INF) printf("impossible\n");
        else printf("%d\n", dp[0][1]);
    }
}