DP[ i ][ j ] 在 i 时刻 j 号车站的等待最小时间.....
有3种可能: 在原地等,坐开往左边的车,做开往右边的车
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A Spy in the Metro
Description
Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. After several thrilling events we find her in the first station of Algorithms City Metro, examining the time table. Maria has an appointment with a local spy at the last station of Algorithms City Metro. Maria knows that a powerful organization is after her. She also knows that while waiting at a station, she is at great The Algorithms City Metro system has N stations, consecutively numbered from 1 to N. Trains move in both directions: from the first station
InputThe input file contains several test cases. Each test case consists of seven lines with information as follows.
The last case is followed by a line containing a single zero. OutputFor each test case, print a line containing the case number (starting with 1) and an integer representing the total waiting time in the stations for a best schedule, or the word `impossible‘ in case Maria is unable to make the appointment. Use the Sample Input4 55 5 10 15 4 0 5 10 20 4 0 5 10 15 4 18 1 2 3 5 0 3 6 10 12 6 0 3 5 7 12 15 2 30 20 1 20 7 1 3 5 7 11 13 17 0 Sample OutputCase Number 1: 5 Case Number 2: 0 Case Number 3: impossible Source Root :: ACM-ICPC World Finals :: 2003 - Beverly Hills Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 9. Dynamic Programming :: Examples |
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N#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=440;
int n,T;
int t[maxn];
int m1,d1[maxn];
int m2,d2[maxn];
int dp[maxn][maxn];
bool hasR[maxn][maxn],hasL[maxn][maxn];
void init()
{
memset(hasR,0,sizeof(hasR));
memset(hasL,0,sizeof(hasL));
memset(dp,63,sizeof(dp));
memset(t,0,sizeof(t));
memset(d1,0,sizeof(d1));
memset(d2,0,sizeof(d2));
}
void getH()
{
for(int i=0;i<m1;i++)
{
int time=d1[i];
for(int j=1;j<=n;j++)
{
time+=t[j];
hasR[time][j]=true;
}
}
for(int i=0;i<m2;i++)
{
int time=d2[i];
for(int j=n;j>=1;j--)
{
time+=t[j+1];
hasL[time][j]=true;
}
}
}
int main()
{
int cas=1;
while(scanf("%d",&n)!=EOF&&n)
{
init();
scanf("%d",&T);
for(int i=2;i<=n;i++)
scanf("%d",t+i);
scanf("%d",&m1);
for(int i=0;i<m1;i++)
scanf("%d",d1+i);
scanf("%d",&m2);
for(int i=0;i<m2;i++)
scanf("%d",d2+i);
getH();
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&&i+t[j+1]<=T&&hasR[i][j])
dp[i][j]=min(dp[i][j],dp[i+t[j+1]][j+1]);
if(j>1&&i+t[j]<=T&&hasL[i][j])
dp[i][j]=min(dp[i][j],dp[i+t[j]][j-1]);
}
}
printf("Case Number %d: ",cas++);
if(dp[0][1]>=INF) puts("impossible");
else printf("%d\n",dp[0][1]);
}
return 0;
}