「POJ3311」Hie with the Pie

题目链接 >http://poj.org/problem?id=3311<

题意：从0出发，经过所有点（点可以重复走）后回到0点，问最短路

思路分析：

　　这题和普通的最短路不太一样，因为题目要求每个点都要走一遍。

　　因此我们选择状压。

　　用SPFA直接开始做，f[i][status]表示到达点i时，状态为status时的最短路，答案就是f[0][(1<<(n+1))-1]

　　更新也和SPFA一模一样

/*By QiXingzhi*/
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define  r  read()
#define  Max(a,b)  (((a)>(b)) ? (a) : (b))
#define  Min(a,b)  (((a)<(b)) ? (a) : (b))
using namespace std;
typedef long long ll;
const int N = 20;
const int INF = 1061109567;
inline int read(){
    int x = 0; int w = 1; register int c = getchar();
    while(c ^ '-' && (c < '0' || c > '9')) c = getchar();
    if(c == '-') w = -1, c = getchar();
    while(c >= '0' && c <= '9') x = (x << 3) +(x << 1) + c - '0', c = getchar();
    return x * w;
}
struct Pizza{
    int u, sta;
};
int n;
int G[N][N],d[N][32768];
queue <Pizza> q;
inline void Debug(int status){
    int tmp[30];
    memset(tmp,0,sizeof(tmp));
    int tot = 0;
    while(status > 0){
        tmp[++tot] = status % 2;
        status /= 2;
    }
    for(int i = tot; i > 0; --i){
        printf("%d",tmp[i]);
    }
}
inline void SPFA(){
//    printf("u = %d  status = ",u); Debug(status); printf("   d = %d
",d[u][status]);
    memset(d,0x3f,sizeof(d));
    d[0][0] = 0;
    Pizza tmp; tmp.u = 0, tmp.sta = 0;
    q.push(tmp);
    int u,status;
    while(!q.empty()){
        Pizza __t = q.front();
        Pizza __next;
        u = __t.u;
        status = __t.sta;
        q.pop();
        for(int i = 0; i <= n; ++i){
            if(G[u][i] == 0) continue;
            if(d[u][status] + G[u][i] < d[i][status ^ (1 << (i))]){
                d[i][status ^ (1 << (i))] = d[u][status] + G[u][i];
                __next.u = i;
                __next.sta = status ^ (1 << (i));
                q.push(__next);
            }
        }
    }
    
}
int main(){
//    freopen(".in","r",stdin);
//    freopen("debug.txt","w",stdout);
    while(1){
        n = r;
        if(n == 0) break;
        for(int i = 0; i <= n; ++i){
            for(int j = 0; j <= n; ++j){
                G[i][j] = r;
            }
        }
        SPFA();
        printf("%d
", d[0][(1<<(n+1))-1]);
    }
    return 0;
}