Tinkoff Challenge - Elimination Round D.Presents in Bankopolis（区间DP）

传送门

题意

给出n条有项边，且点都在一条直线上从1到n编号，询问走过k-1条边的最短距离，注意，走的过程中不能经过先前访问过的点，比如1->3->5->2不行，在到达2
的时候经过了已访问点3

分析

区间DP，dp[l][r][dir][step]表示l到r方向为dir步数为step的最短距离，我们呢可以递归求解，dp[l][r][dir][step]由dp[l][i][1][step+1]+c[r][i]和dp[i][r][0][step+1]+c[l][i]转移过来,一开始dp[0][n+1][dir][0]

Oleg can't go near offices he visited, so if now he is in office v, he visited q, q > v, he can't visit office k, where k > q. That means if now he's in office v, he can only visit office q, l < v < r, l is rightmost visited office which position is less than v and r is leftmost visited office which position is more than v. So we have to solve our problem for state [v][l][r][s], where s count of steps we made. We have 2 options. First is go to left and second is go to right. If we go left, we are not interested what is r, and if we go right we are not interested what is l. So our state now is [left border][right border][direction][s]. State [v][l][r][s] becomes 2: [v][r][right][s] and [l][v][left][s]. We can exactly identify our current position. If we go left it's right border, and if we go right it's left border. Now, when we are in state [l][r][direction][s] and now we are in office v, we have to just find all offices q, l<=q<=r and c[v][q] != 0 and solve our problem for [l][q][right][s+1] and [q][r][left][s+1] and take minimum. For it we can use recursion. We can start from any office, so we have to solve problem for [0][v][left][0] and [v][n+1][right][0] and take minimum.
We have n^22k states. In each state we check make not more than n operations. So algo works in O(n^3*k)

trick

代码

#include<bits/stdc++.h>
using namespace std;

int dp[100][100][2][100];
int vis[100][100][2][100];
int c[100][100];
int n,k,m,u,v,value;
int work(int l,int r,int dir,int step)
{
    if(step==(k-1)) return 0;
    if(vis[l][r][dir][step]) return dp[l][r][dir][step];
    vis[l][r][dir][step]=1;
    int bit=dir?r:l;int ret=1e9;
    for(int i=l+1;i<r;++i) if(c[bit][i])
    {
        ret=min(ret,c[bit][i]+work(i,r,0,step+1));
        ret=min(ret,c[bit][i]+work(l,i,1,step+1));
    }
    dp[l][r][dir][step]=ret;
    return ret;
}


int main()
{
    scanf("%d %d",&n,&k);
    scanf("%d",&m);
    for(int i=1;i<=m;++i)
    {
        scanf("%d %d %d",&u,&v,&value);
        if(c[u][v]==0) c[u][v]=value;
        else c[u][v]=min(c[u][v],value);
    }
    int ans=1e9;
    for(int i=1;i<=n;++i)
    {
        ans=min(ans,work(0,i,1,0));
        ans=min(ans,work(i,n+1,0,0));
    }
    if(ans==1e9) puts("-1");else printf("%d\n",ans);
}