【OpenJudge 4124】海贼王之伟大航路（DP DFS）

海贼王之伟大航路

 OpenJ_Bailian - 4124 

两种解法：

第一种用DFS，用一个record数组保存每个状态最优解，大于这个值就剪枝不搜了。

#include<bits/stdc++.h>#define INF 0X3f3f3f3f#define LL long longusing namespace std;int n,a[20][20],ans,book[20],step,record[20][1<<15],weight[20],pos;//record[i][j]表示通过j状态到达i岛时的最优路径，经过一个岛就加上那个岛的weight值。pos表示当前状态。void dfs(int cur,int now) //cur是已经去过的岛的数目，now是现在在哪个岛{    if(cur+2 == n)    {        step += a[now][n];        if(step < ans)  ans=step;        step -= a[now][n];        return;    }    if(step>ans)        return;    for(int i = 2; i < n; ++i)    {        if(!book[i])        {            if(record[i][pos+weight[i]] <= step+a[now][i])                continue;            step += a[now][i];            pos += weight[i];            book[i] = 1;            record[i][pos] = step;            dfs(cur+1,i);            step -= a[now][i];            pos -= weight[i];            book[i] = 0;        }    }    return;}int main(){    for(int i = 1; i <= 16; ++i)        weight[i] = 1 << i;    while(~scanf("%d",&n))    {        pos = step = 0; ans = INF;        memset(book,0,sizeof book);        memset(record,INF,sizeof record);        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= n; ++j)                scanf("%d",&a[i][j]);        dfs(0,1);        printf("%d\n",ans);    }    return 0;}