UVa 1025 A Spy in the Metro

最新推荐文章于 2021-05-21 21:43:19 发布

Kim0403

最新推荐文章于 2021-05-21 21:43:19 发布

阅读量344

点赞数

CC 4.0 BY-SA版权

分类专栏： UVa 动态规划-基础入门经典-竞赛篇

本文链接：https://blog.youkuaiyun.com/Kim0403/article/details/51833081

UVa 同时被 3 个专栏收录

9 篇文章

订阅专栏

动态规划-基础

9 篇文章

订阅专栏

入门经典-竞赛篇

8 篇文章

订阅专栏

本文介绍了一个列车调度算法的具体实现，该算法通过动态规划方法计算出在给定时间限制内完成从起点到终点行程所需的最少列车数量。算法考虑了不同站点间的行驶时间和多个列车的到达时刻，确保列车调度的有效性和可行性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<deque>
#include<functional>
#include<iterator>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#define CPY(A, B) memcpy(A, B, sizeof(A))
typedef long long LL;
typedef unsigned long long uLL;
const int MOD = int (1e9) + 7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-9;
const double OO = 1e20;
const double PI = acos (-1.0);
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
int dp[210][55];
int has_train[210][55][2];
int t[210],d1[55],d2[55];
int main() {
    int n,T,m1,m2; int kase=1;
    while (~scanf ("%d",&n) &&n) {
        cin>>T;
        for (int i=1; i<n; ++i) { cin>>t[i]; }
        cin>>m1;
        for (int i=1; i<=m1; ++i) { cin>>d1[i]; }
        cin>>m2;
        for (int i=1; i<=m2; ++i) { cin>>d2[i]; }
        memset (has_train,0,sizeof (has_train) );
        for (int i=1; i<=m1; ++i) {
            has_train[d1[i]] [1][0]=1;//ith train when go from station 1
            int temp=d1[i];
            for (int j=1; j<n; j++) {
                temp+=t[j];//station one by one
                if (temp<=T) { has_train[temp][j+1][0]=1; }//when arrive station j
                else { break; }
            }
        }
        for (int i=1; i<=m2; i++) {
            has_train[d2[i]] [n][1]=1;
            int temp=d2[i];
            for (int j=n-1; j>1; j--) {
                temp+=t[j];
                if (temp<=T) { has_train[temp][j][1]=1; }
                else { break; }
            }
        }
        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]);
                }
            }
        cout<<"Case Number "<<kase++<<": ";
        if (dp[0][1]>=INF) { cout<<"impossible\n"; }
        else { cout<<dp[0][1]<<"\n"; }
    }
    return 0;
}