poj3311 TSP问题 状压DP

题意：从0走遍1~n最后再返回到0，一个点可以走多次，求经过的最短距离。

分析：由于一个点可以走多次，所以需要求出任意两点间的最短距离，就要用到floyd算法，同时接下来可以搜索做复杂度是O(n!),而状态压缩的时间效率就高了，关于DP自己有时候状态转移方程知道了，却不知道如何去实现，循环时哪层放在外面，哪层放在里面，所以很容易把DP写成了暴力的搜索，自己要多想想。

状态转移方程：dp[S][i] = min(dp[S^(1<<(i-1))][j] + dis[j][i],dp[S][i])

四不像的搜索：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <cctype>#include <vector>#include <set>#include <map>#include <queue>using namespace std;#define ll long longconst int maxn = (1<<10) + 5;const int inf = 0x7f7f7f7f;int g[15][15];int n;int mmin;inline int getone(int x){    int res = 0;    while (x)    {        x &= (x-1);        res++;    }    return res;}void floyd(){    for (int k = 0; k <= n; k++)    {        for (int i = 0; i <= n; i++)        {            for (int j = 0; j <= n; j++)                g[i][j] = min(g[i][j],g[i][k]+g[k][j]);        }    }}void dfs(int u, int state, int cur)//写搜索本可以不用二进制{    if (getone(state)==0)    {        if (mmin > cur+g[u][0]) mmin = cur+g[u][0];        return;    }    for (int i = 1; i <= n; i++)    {        if (state&(1<<(i-1)) && g[u][i] != -1)        {            dfs(i, (1<<(i-1))^state, cur+g[u][i]);        }    }}int main(){    //freopen("input.txt", "r", stdin);    while (cin >> n && n)    {       for (int i = 0; i <= n; i++)            for (int j = 0; j <= n; j++)cin >> g[i][j];        floyd();        mmin = inf;        for (int i = 1; i <=n; i++)        {            dfs(i, (1<<(i-1))^((1<<n)-1), g[0][i]);        }        cout << mmin << endl;    }}

DP：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <cctype>#include <vector>#include <set>#include <map>#include <queue>using namespace std;#define ll long longconst int maxn = (1<<10) + 5;const int inf = 0x7f7f7f7f;int g[15][15];int f[maxn][12];int n;int mmin;void floyd(){    for (int k = 0; k <= n; k++)    {        for (int i = 0; i <= n; i++)        {            for (int j = 0; j <= n; j++)                g[i][j] = min(g[i][j],g[i][k]+g[k][j]);        }    }}int main(){    //freopen("input.txt", "r", stdin);    while (cin >> n && n)    {        for (int i = 0; i <= n; i++)            for (int j = 0; j <= n; j++)cin >> g[i][j];        floyd();        for (int s = 1; s < (1<<n); s++)//要先枚举状态，跟floyd有点像，此时i，j之间可以进行松弛操作        {            for (int i = 1; i <= n; i++)            {                if (s&(1<<(i-1)))                {                    if (s == (1<<(i-1)))f[s][i] = g[0][i];                    else                    {                        f[s][i] = inf;                        for (int j = 1; j <= n; j++)                            if (s&(1<<(j-1)) && i != j)                                f[s][i] = min(f[s^(1<<(i-1))][j]+g[j][i], f[s][i]);                    }                }            }        }        int ans = inf;        for (int i = 1; i <= n; i++)            if (ans > f[(1<<n)-1][i]+g[i][0]) ans = f[(1<<n)-1][i]+g[i][0];        cout << ans << endl;    }}