UVA1025-A Spy in the Metro(动态规划)

Problem UVA1025-A Spy in the Metro

Accept: 713  Submit: 6160
Time Limit: 3000 mSec

Problem Description

Input

 Output

For 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 format of the sample output.

 

 Sample Input

4
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 Output

Case Number 1: 5

Case Number 2: 0

Case Number 3: impossible

题解：很明显的动态规划，dp[i][j]表示i时刻在j站点还需要的最短等待时间，总共就三种选择，状态转移方程很简单，边界一直是我写动态规划题比较头疼的地方，不过这个题还比较简单，t时刻在n站点自然是0，在其他站点就是INF（为了不会从这些状态转移过去）。

#include <bits/stdc++.h>

using namespace std;

const int maxt = 200 + 10, maxn = 50 + 5;
const int INF = 0x3f3f3f3f;

int read() {
    int q = 0; char ch = ' ';
    while (ch<'0' || ch>'9') ch = getchar();
    while ('0' <= ch && ch <= '9') {
        q = q * 10 + ch - '0';
        ch = getchar();
    }
    return q;
}

int n, t, m1, m2;
int ti[maxn];
int dp[maxt][maxn];
bool have_train[maxt][maxn][2];

void init() {
    memset(have_train, false, sizeof(have_train));
    for (int i = 1; i <= n - 1; i++) {
        dp[t][i] = INF;
    }
    dp[t][n] = 0;
}

int T = 1;

int main()
{
    //freopen("input.txt", "r", stdin);
    while (scanf("%d", &n) && n) {
        t = read();
        init();
        for (int i = 1; i < n; i++) {
            ti[i] = read();
        }
        ti[n] = ti[0] = INF;
        m1 = read();
        int d;
        for (int i = 0; i < m1; i++) {
            d = read();
            int cnt = 1;
            for (int j = 1; j <= n; j++) {
                have_train[d][cnt][0] = true;
                //printf("d:%d cnt:%d
", d, cnt);
                cnt++;
                d += ti[j];
            }
        }
        //printf("
");

        m2 = read();
        for (int i = 0; i < m2; i++) {
            d = read();
            int cnt = n;
            for (int j = n; j >= 1; j--) {
                have_train[d][cnt][1] = true;
                //printf("d:%d cnt:%d
", d, cnt);
                cnt--;
                d += ti[j - 1];
            }
        }

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

                if (i + ti[j - 1] <= t && have_train[i][j][1]) {
                    dp[i][j] = min(dp[i][j], dp[i + ti[j - 1]][j - 1]);
                }
            }
        }

        printf("Case Number %d: ", T++);
        if (dp[0][1] >= INF) {
            printf("impossible
");
        }
        else {
            printf("%d
", dp[0][1]);
        }
    }
    return 0;
}