HDU 3001 Travelling (三进制状态压缩DP+BFS)

题意：有n个点（n<=10），要求都经过，每个点最多经过两次，求遍历每个点的最小花费。

分析：每个点最多经过两次，则用三进制保存状态，一个好的技巧就是开个pos数组，记录每个状态用三进制表示时，每一位的情况。    枚举每个点为起点，然后每次sum值是不需要清空的，清空了反而造成了诸多重复遍历....................

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#define INF 0x7FFFFFFFusing namespace std;int p[11];int pos[60000][11];int dist[11][11],sum[11][60000];struct node {    int x,buff;} q[1111111];int head,tail,n,m,ans;void init() {    for(int i=1; i<=10; i++)        for(int j=1; j<=10; j++)            if(i != j) dist[i][j] = INF;            else dist[i][j] = 0;    memset(sum,0,sizeof(sum));}void bfs(int v0) {    ans = INF;    // memset(sum,0,sizeof(sum));    head = 0;    tail = 0;    q[head].x = v0;    q[head++].buff = p[v0-1];    while(head != tail) {        node t = q[tail ++];        node tt;        int cnt = 0;        for(int i=0; i<=n; i++) {            if(pos[t.buff][i] >= 1) cnt++;        }        if(cnt == n) {            ans = min(ans,sum[t.x][t.buff]);        }        for(int i=1; i<=n; i++) {            if(t.x == i || dist[t.x][i] == INF) continue;            if(pos[t.buff][i-1] >= 2) continue;            if(sum[t.x][t.buff] + dist[t.x][i] >= ans) continue;            tt.buff = t.buff + p[i-1];            tt.x = i;            if(sum[i][tt.buff] != 0 && sum[i][tt.buff] > sum[t.x][t.buff] + dist[t.x][i]) {                sum[i][tt.buff] = sum[t.x][t.buff] + dist[t.x][i];                q[head++] = tt;            } else if(! sum[i][tt.buff]) {                sum[i][tt.buff] = sum[t.x][t.buff] + dist[t.x][i];                q[head++] = tt;            }        }    }}int main() {    int a,b,c;    for(int i=1; i<=11; i++) p[i-1] = pow(3.0,(i-1)*1.0);    for(int i=1; i<=59050; i++) {        int tmp = i,j = 0;        while(tmp) {            pos[i][j++] = tmp % 3;            tmp = tmp / 3;        }    }    while(scanf("%d%d",&n,&m) != EOF) {        init();        for(int i=0; i<m; i++) {            scanf("%d%d%d",&a,&b,&c);            if(c < dist[a][b]) {                dist[a][b] = c;                dist[b][a] = c;            }        }        int minn = INF;        for(int i=1; i<=n; i++) {            bfs(i);            minn = min(minn,ans);        }        if(minn == INF) printf("-1\n");        else printf("%d\n",minn);    }    return 0;}