题目链接:http://poj.org/problem?id=3311
题目:
题意:n个城市,每两个城市间都存在距离,问你恰好经过所有城市一遍,最后回到起点(0)的最短距离。
思路:我们首先用floyd预处理出每两个城市间的最短路,然后采用状压dp来解题。dp[i][j]表示在i这种状压下以j为目标城市的最短距离,i的二进制中x位为1表示到了城市x,为0表示没到城市x,则转移方程为dp[i][j] = min(dp[i][j], dp[i^(1<<(j-1))][k] + dis[k][j]),其中i^(1<<(j-1))表示没到城市j。
代码实现如下:
1 #include <set> 2 #include <map> 3 #include <queue> 4 #include <stack> 5 #include <cmath> 6 #include <bitset> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<ll, ll> pll; 18 typedef pair<ll, int> pli; 19 typedef pair<int, ll> pil;; 20 typedef pair<int, int> pii; 21 typedef unsigned long long ull; 22 23 #define lson i<<1 24 #define rson i<<1|1 25 #define bug printf("********* "); 26 #define FIN freopen("D://code//in.txt", "r", stdin); 27 #define debug(x) cout<<"["<<x<<"]" <<endl; 28 #define IO ios::sync_with_stdio(false),cin.tie(0); 29 30 const double eps = 1e-8; 31 const int mod = 10007; 32 const int maxn = 1e6 + 7; 33 const double pi = acos(-1); 34 const int inf = 0x3f3f3f3f; 35 const ll INF = 0x3f3f3f3f3f3f3f; 36 37 int n; 38 int dp[2050][15], dis[15][15]; 39 40 void floyd() { 41 for(int k = 0; k <= n; k++) { 42 for(int i = 0; i <= n; i++) { 43 for(int j = 0; j <= n; j++) { 44 dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); 45 } 46 } 47 } 48 } 49 50 int main() { 51 //FIN; 52 while(~scanf("%d", &n) && n) { 53 for(int i = 0; i <= n; i++) { 54 for(int j = 0; j <= n; j++) { 55 scanf("%d", &dis[i][j]); 56 } 57 } 58 floyd(); 59 memset(dp, -1, sizeof(dp)); 60 dp[1][0] = 0; 61 int tot = 1 << (n+1); 62 for(int i = 0; i < tot; i++) { 63 for(int j = 1; j <= n; j++) { 64 if(i & (1 << (j-1))) { 65 if(i == (1<<(j-1))) dp[i][j] = dis[0][j]; 66 else { 67 dp[i][j] = inf; 68 for(int k = 1; k <= n; k++) { 69 if(i & (1 << (k-1)) && j != k) { 70 dp[i][j] = min(dp[i][j], dp[i^(1<<(j-1))][k] + dis[k][j]); 71 } 72 } 73 } 74 } 75 } 76 } 77 int ans = inf; 78 for(int i = 1; i <= n; i++) { 79 ans = min(ans, dp[(1<<n)-1][i] + dis[i][0]); //最后要从某一座城市回到起点 80 } 81 printf("%d ", ans); 82 } 83 return 0; 84 }