MYF

UVa 1025 A Spy in the Metro

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题目链接

UVa 1025

方法：DP

题目分析

题目大意

见紫书P267

解析

用dp(i, j)记录在i时刻在第j个地铁站时所需要等待的最少时间，所以dp(0, 1)表示在零时刻所需等待的时间。将i从后往前递推，可以求出来dp(0, 1)，即为解。

代码

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#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <deque>
#include <algorithm>
#define Memset(a,val) memset(a,val,sizeof(a))
#define PI acos(-1)
#define rt(n)       (i == n ? '\n' : ' ')
#define hi         printf("Hi----------\n")
#define IN freopen("input.txt","r",stdin);
#define OUT freopen("output.txt","w",stdout);
#define debug(x) cout<<"Debug : ---"<<x<<"---"<<endl;
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long ll;
using namespace std;
const int maxn=100000+5;
const int mod=1000000007;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int main(){
    int kase=1;
    int n,T,t[220],f[220],e[220],dp[220][70],has_train[220][70][2];
    while (~scanf("%d",&n)&&n) {
        scanf("%d",&T);
        for (int i=1; i<n; i++) {
            scanf("%d",&t[i]);
        }
        int m1,m2;
        scanf("%d",&m1);
        for (int i=1; i<=m1; i++) {
            scanf("%d",&f[i]);
        }
        scanf("%d",&m2);
        for (int i=1; i<=m2; i++) {
            scanf("%d",&e[i]);
        }
        Memset(has_train, 0);
        
        for (int i=1; i<=m1; i++) {
            int tt=f[i];
            has_train[tt][1][0]=1;
            for (int j=1; j<n; j++) {
                tt+=t[j];
                if (tt>T) break;
                has_train[tt][j+1][0]=1;
            }
        }
        
        for (int i=1; i<=m2; i++) {
            int tt=e[i];
            has_train[tt][n][1]=1;
            for (int j=n; j>1; j--) {
                tt+=t[j-1];
                if (tt>T) break;
                has_train[tt][j-1][1]=1;
            }
        }
        
        for (int i=1; i<=n; 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]);
                }
            }
        }
        cout<<"Case Number "<<kase++<<": ";
        if (dp[0][1]>=INF) 
            cout<<"impossible\n";
        else
            cout<<dp[0][1]<<endl;
    }
}