2011 成都 D Disney's Fastpass

原创于 2019-07-18 07:42:03 发布 · 252 阅读

CC 4.0 BY-SA版权

本文提供了一道 ACM 竞赛编程题的详细解答过程，采用动态规划与最短路径算法解决复杂的任务分配与路径寻找问题。通过具体实例展示了输入输出样例及核心代码实现。

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=520

#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 50;
const int MAX_K = 8;
const int INF = INT_MAX / 10;
int g[MAX_N][MAX_N];
int dp[MAX_N][1 << MAX_K][1 << MAX_K];
int pos[MAX_K], t[MAX_K], ft[MAX_K];
int p[MAX_N];
int solve()
{
    int N, M, K;
    scanf("%d%d%d", &N, &M, &K);
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            g[i][j] = INF;
    for (int i = 0; i < N; i++)
        g[i][i] = 0;
    for (int i = 0; i < N; i++)
        for (int j = 0; j < (1 << K); j++)
            for (int k = 0; k < (1 << K); k++)
                dp[i][j][k] = INF;
    dp[0][0][0] = 0;
    for (int i = 0; i < M; i++) {
        int a, b, d;
        scanf("%d%d%d", &a, &b, &d);
        a--, b--;
        g[a][b] = g[b][a] = d;
    }
    for (int k = 0; k < N; k++)
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
    for (int i = 0; i < N; i++)
        p[i] = 0;
    for (int i = 0; i < K; i++) {
        int num;
        scanf("%d%d%d%d", &pos[i], &t[i], &ft[i], &num);
        pos[i]--;
        for (int j = 0; j < num; j++) {
            int t;
            scanf("%d", &t);
            t--;
            p[t] |= (1 << i);
        }
    }
    int ans = INF;
    for (int s1 = 0; s1 < (1 << K); s1++)
        for (int s2 = 0; s2 < (1 << K); s2++)
            for (int i = 0; i < N; i++) {
                if (dp[i][s1][s2] == INF)
                    continue;
                if (s2 == ((1 << K) - 1))
                    ans = min(ans, dp[i][s1][s2] + g[i][0]);
                for (int j = 0; j < K; j++) {
                    if (s2 & (1 << j))
                        continue;
                    int &r = dp[pos[j]][s1 | p[pos[j]]][s2 | (1 << j)];
                    r = min(r, dp[i][s1][s2] + (s1 & (1 << j) ? ft[j] : t[j]) + g[i][pos[j]]);
                }
                for (int j = 0; j < N; j++) {
                    int &r = dp[j][s1 | p[j]][s2];
                    r = min(r, dp[i][s1][s2] + g[i][j]);
                }
            }
    return ans;
}
int main()
{
    int T;
    scanf("%d", &T);
    for (int i = 1; i <= T; i++)
        printf("Case #%d: %d\n", i, solve());
}