hdu 5092 线裁剪(纵向连线最小和+输出路径)

http://acm.hdu.edu.cn/showproblem.php?pid=5092

给一个m*n的矩阵，找到一个纵向的"线"使得线上的和最小并输出这条线，线能向8个方向延伸，要求找的是纵向的一条线(每一行各取一个点连成一线)

比较裸的dp，当前点只受到其上一行中的三个点的影响，然后求一下最大连和即可，dp过程中记录路径，然后打印时回溯即可

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int inf = 0x3fffffff;
const LL _inf = 1e18;
const int maxn = 105,maxm = 10005;
int p[maxn][maxn],dp[maxn][maxn],col[maxn][maxn],n,m,cas = 1;
char ch[maxn];
bool in(int x,int y)
{
    return 1<=x&&x<=m&&1<=y&&y<=n;
}
void print(int x,int y)
{
    if(x == 1){
        printf("%d",y);
        return ;
    }
    print(x - 1,col[x][y]);
    printf(" %d",y);
}
void work()
{
    RD2(m,n);
    for(int i = 1;i <= m;++i)
        for(int j = 1;j <= n;++j){
            RD(p[i][j]);
            dp[i][j] = inf;
        }
    for(int j = 1;j <= n;++j)
        dp[1][j] = p[1][j];

    for(int i = 2;i <= m;++i)
    for(int j = n;j >= 1;--j){
        for(int k = j+1;k >= j-1;--k){
            if(in(i-1,k) && dp[i][j] > dp[i-1][k] + p[i][j]){
                dp[i][j] = dp[i-1][k] + p[i][j];
                col[i][j] = k;
            }
        }
    }
    int mn = inf;
    for(int j = n;j >= 1;--j)
        mn = min(mn,dp[m][j]);
    //printf("%d
",mn);
    printf("Case %d
",cas++);
    for(int j = n;j >= 1;--j)
    if(mn == dp[m][j]){
        print(m,j);
        puts("");
        return ;
    }
}
int main()
{
    int _;RD(_);
    while(_--){
        work();
    }
    return 0;
}