POJ2686 Traveling by Stagecoach 状态压缩DP

POJ2686

比较简单的 状态压缩DP

注意DP方程转移时，新的状态必然数值上小于当前状态，故最外层循环为状态从大到小即可。

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1 << 10;
const int maxm = 31;
const int INF = 1 << 29;  

int n, m, p, a, b;
int t[maxm];
int d[maxm][maxm];   //图的邻接矩阵表示（-1表示没有边）  

//dp[S][v] := 到达剩下的车票集合为S并且现在在城市v的状态所需要的最小花费
double dp[maxn][maxm];  

void solve()
{
    for (int i = 0; i < (1 << n); i++)
        fill(dp[i], dp[i] + m + 1, INF);    //用足够大的值初始化  

    dp[(1 << n) - 1][a] = 0;
    double res = INF;
    for (int i = (1 << n) - 1; i >= 0; i--){
        for (int u = 1; u <= m; u++){
            for (int j = 0; j < n; j++){
                if (i & (1 << j)){
                    for (int v = 1; v <= m; v++){
                        if (d[v][u]){
                            //使用车票i，从v移动到u
                            dp[i & ~(1 << j)][v] = min(dp[i & ~(1 << j)][v], dp[i][u] + (double)d[u][v] / t[j]);
                        }
                    }
                }
            }
        }
    }
    for (int i = 0; i < (1 << n); i++)
        res = min(res, dp[i][b]);
    if (res == INF)
        //无法到达
        printf("Impossible\n");
    else
        printf("%.3f\n", res);  

}  

int main()
{
    while(scanf("%d%d%d%d%d", &n, &m, &p, &a, &b)!= EOF){
        if(!n && !m)
        break;
        memset(d, 0, sizeof(d));
        for (int i = 0; i < n; i++)
            scanf("%d", &t[i]);
        for (int i = 0; i < p; i++){
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            d[u][v] = d[v][u] = c;
        }
        solve();
    }
    return 0;
}