紫书动规 例题9-4 UVA - 116 Unidirectional TSP 多段图的最短路 dp

题目链接:

https://vjudge.net/problem/UVA-116

题意:

题解:

dp[i][j]:= 从(i,j)出发到最后一列的最小开销
因为字典序最小，所以每次往前一列转移，都要先从这列 行数最小的位置转移
dp[i][j] = min(dp[i][j],dp[row[k]][j+1]+a[i][j]);
边界是dp[i][m-1] = a[i][m-1]

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

int n,m;
int a[15][105],dp[15][105];
int first,nx[15][105];

int main(){
    while(cin >> n >> m){
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++){
                cin >> a[i][j];
            }
        memset(dp,0x3f,sizeof(dp));
        for(int i=0; i<n; i++) dp[i][m-1] = a[i][m-1];

        int ans = INF;
        for(int j=m-1; j>=0; j--){
            for(int i=0; i<n; i++){
                int row[3] = {i,i-1,i+1};
                if(row[1]<0) row[1]=n-1;
                if(row[2]==n) row[2]=0;
                sort(row,row+3);
                for(int k=0; k<3; k++){
                    int t = dp[i][j];
                    dp[i][j] = min(dp[i][j],dp[row[k]][j+1]+a[i][j]);
                    if(t != dp[i][j]) { nx[i][j] = row[k]; }
                }
                if(j == 0 && dp[i][j]<ans) { ans=dp[i][j]; first=i; };
            }
        }

        printf("%d",first+1);
        for(int i=nx[first][0],j=1; j<m; i=nx[i][j],j++){
            printf(" %d",i+1);
        }
        printf("\n%d\n",ans);
    }

    return 0;
}