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POJ 2686 Traveling by Stagecoach (状压DP)
题意:有一个人从某个城市要到另一个城市, 有n个马车票,相邻的两个城市走的话要消耗掉一个马车票。花费的时间呢,是马车票上有个速率值
,问最后这个人花费的最短时间是多少。
析:和TSP问题差不多,dp[s][i] 表示当前在第 i 个城市,还剩余集合 s的票,需要的最短时间。状态转移方程:
dp[s][i] = min{dp[s|j][k] + d[i][k] }
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e6 + 10; const int mod = 100000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } double dp[1<<8][31]; int a[31][31]; double tt[10]; int main(){ // freopenr; int s, t, p; while(scanf("%d %d %d %d %d", &n, &m, &p, &s, &t) == 5 && n+m+p+s+t){ int all = 1 << n; for(int i = 0; i < all; ++i) fill(dp[i], dp[i]+m+1, inf); for(int i = 0; i < n; ++i) scanf("%lf", tt+i); memset(a, INF, sizeof a); for(int i = 0; i < p; ++i){ int u, v, d; scanf("%d %d %d", &u, &v, &d); a[u][v] = a[v][u] = d; } dp[all-1][s] = 0; for(int i = all-2; i >= 0; --i) for(int j = 0; j < n; ++j) if(!(i&(1<<j))){ for(int u = 1; u <= m; ++u){ for(int v = 1; v <= m; ++v){ if(a[u][v] == INF) continue; if(dp[i|(1<<j)][v] == inf) continue; dp[i][u] = min(dp[i][u], dp[i|(1<<j)][v] + a[u][v] * 1.0 / tt[j]); } } } double ans = inf; for(int i = 0; i < all; ++i) ans = min(ans, dp[i][t]); if(ans >= inf) printf("Impossible\n"); else printf("%.3f\n", ans); } return 0; }
POJ 2686 Traveling by Stagecoach (状压DP)
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