UVa1025 A Spy in the Metro(dp)

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分析：
紫书 P267
AC代码：
#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;int a[500];//   起点发车时刻int b[500];//   终点发车时刻int Time[500];//  车站到车站之间的距离int dp[500][500];int judge[500][500][10];int main (){    int n,T,M1,Me;    int cc=1;    //FIN;    while (scanf ("%d",&n)&&n){        mset(Time,0);mset(judge,0);        scanf ("%d",&T);        for (int i=1;i<n;i++) scanf ("%d",&Time[i]);        scanf ("%d",&M1);//     起点        for (int i=1;i<=M1;i++) scanf ("%d",&a[i]);        scanf ("%d",&Me);//     终点        for (int i=1;i<=Me;i++) scanf ("%d",&b[i]);        for (int i=1;i<=M1;i++){//   起点到终点有车的车站时刻            int tt=a[i];            judge[tt][1][0]=1;            for (int j=1;j<n;j++){                judge[tt+=Time[j]][j+1][0]=1;            }        }        for (int i=1;i<=Me;i++){//   终点到起点有车的车站时刻            int tt=b[i];            judge[tt][n][1]=1;            for (int j=n-1;j>=1;j--){                judge[tt+=Time[j]][j][1]=1;            }        }        for (int i=0;i<=T;i++)            for (int j=0;j<=n;j++) dp[i][j]=INF;        for (int i=0;i<=n;i++) dp[i][n+1]=dp[i][0]=0;        dp[0][1]=0;        for (int i=1;i<=T;i++){            for (int j=1;j<=n;j++){                dp[i][j]=dp[i-1][j]+1;                if(j>1&&judge[i][j][0]&&i>=Time[j-1])//   从左向右                    dp[i][j]=min(dp[i][j],dp[i-Time[j-1]][j-1]);                if(j<n&&judge[i][j][1]&&i>=Time[j])//    从右向左                    dp[i][j]=min(dp[i][j],dp[i-Time[j]][j+1]);            }        }        printf ("Case Number %d: ",cc++);        if (dp[T][n]>=INF)printf ("impossible\n");        else printf ("%d\n",dp[T][n]);    }    return 0;}
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