poj2686(Traveling by Stagecoach)状态压缩dp

标签：acm   状压dp   

3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0

30.000
3.667
Impossible
Impossible
2.856

30.0 
3.66667 
Impossible 
Impossible 
2.85595

题意：m个城市，p条路，告诉你每条路长度，给你n张车票，每张车票都有一个车速。一条路必须要用掉一张车票，时间为距离/车速。求从a到b所需的最短时间。

题解：状态压缩dp，dp[s][v]表示还剩的车票集合为s，在v城市时最短时间，那么

dp[s][v] = min{ dp[s-{i}][u] + d(v, u) }

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 10;
const int maxm = 35;

double dp[(1<<maxn)-1][maxm];
int d[maxm][maxm];
int t[maxn];
int vis[maxm];
int n, m, p, a, b;

double rec(int s, int v)
{
    vis[v] = 1;
    if (dp[s][v] >= 0)
        return dp[s][v];
    if (v == b)
        return dp[s][v] = 0;
    double res = INF;
    for (int u = 0; u < m; ++u)
        for (int i = 0; i < n; ++i)
            if (u != v && !vis[u] && (1<<i)&s && d[u][v] < INF)
            {
                vis[u] = 1;
                res = min(res, rec((1<<i)^s, u)+d[v][u]*1.0/t[i]);
                vis[u] = 0;
            }
    return dp[s][v] = res;
}

int main()
{
    while (cin >> n >> m >> p >> a >> b && (n || m || p || a || b))
    {
        a--;
        b--;
        for (int i = 0; i < n; ++i)
            scanf("%d", &t[i]);
        fill(d[0], d[m], INF);
        fill(vis, vis+m, 0);
        fill(dp[0], dp[1<<n], -1);
        while (p--)
        {
            int x, y, dis;
            scanf("%d%d%d", &x, &y, &dis);
            d[x-1][y-1] = d[y-1][x-1] = dis;
        }
        double ans = rec((1<<n)-1, a);
        if (fabs(ans-INF) < 1e-8)
            printf("Impossible\n");
        else
            printf("%.3f\n", ans);
    }
    return 0;
}

poj2686(Traveling by Stagecoach)状态压缩dp

标签：acm   状压dp   

原文地址：http://blog.csdn.net/god_weiyang/article/details/47782345