題意： 將所有外賣送去所有對應地點再回到店鋪，求最短路。 分析：Floyed預處理任意兩點最小距離，然後二維狀壓(最後的結束位置要考慮，不然回到起點的距離沒辦法計算)。 $dp[j + 1][(1 << j) + i] = min(dp[j + 1][(1 << j) + i], dp[k + 1][i] + mp[k + 1][j + 1])：表示以第j個位置結束的狀態i的路徑長度$

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
#define met(s) memset(s, 0, sizeof(s))
#define rep(i, a, b) for(int i = a; i <= b; ++i)
template <class T> inline void scan_d(T &ret) {
char c; ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0'), c = getchar();}}
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int mod = 1e8;
const int MAXN = 15;
int dp[MAXN][1 << 12];
int mp[MAXN][MAXN];
int n, m;

inline void floyed() {
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= n; ++j) {
			for(int k = 1; k <= n; ++k) {
				mp[j][k] = min(mp[j][k], mp[j][i] + mp[i][k]);
			}
		}
	}
}

int main() {
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	while(scanf("%d", &n) && n) {
		n += 1;
		memset(dp, INF, sizeof(dp));
		for(int i = 1; i <= n; ++i) {
			for(int j = 1; j <= n; ++j) {
				scanf("%d", &mp[i][j]);
			}
		}
		floyed(); dp[1][1] = 0; int ans = INF;
		//for(int i = 0; i < n; ++i) dp[1 << i] = mp[1][i + 1];
		for(int i = 1; i < (1 << n); ++i) {
			if(!(i & 1)) continue;
			for(int j = 0; j < n; ++j) {
				if((1 << j) & i) continue;
				for(int k = 0; k < n; ++k) {
					if(!((1 << k) & i)) continue;
					dp[j + 1][(1 << j) + i] = min(dp[j + 1][(1 << j) + i], dp[k + 1][i] + mp[k + 1][j + 1]);
					if((1 << n) - 1 == i + (1 << j)) {
						ans = min(ans, dp[j + 1][(1 << n) - 1] + mp[j + 1][1]);
					//	printf("## %d %d\n", mp[k + 1][1], k + 1);
					}
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}
 

$next$_$permutation:$

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
#define met(s) memset(s, 0, sizeof(s))
#define rep(i, a, b) for(int i = a; i <= b; ++i)
template <class T> inline void scan_d(T &ret) {
char c; ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0'), c = getchar();}}
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int mod = 1e8;
const int MAXN = 15;
int a[15];
int mp[MAXN][MAXN];
int n, m;

inline void floyed() {
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= n; ++j) {
			for(int k = 1; k <= n; ++k) {
				mp[j][k] = min(mp[j][k], mp[j][i] + mp[i][k]);
			}
		}
	}
}

int main() {
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	while(scanf("%d", &n) && n) {
		n += 1;
		for(int i = 1; i <= n; ++i) {
			a[i] = i;
			for(int j = 1; j <= n; ++j) {
				scanf("%d", &mp[i][j]);
			}
		}
		floyed(); int ans = INF;
		do {
			int sum = mp[a[n]][1] + mp[1][a[1]];
			for(int i = 1; i < n; ++i) {
				sum += mp[a[i]][a[i + 1]];
			}
			ans = min(sum, ans)
;		}while(next_permutation(a + 2, a + n + 1));
		printf("%d\n", ans);
	}
	return 0;
}