uva116

这题是一道dp的水题,类似于数塔,从右向左倒着加上去,找到最小值,然后在从左到右输出路径。

#include"iostream"
#include"stdio.h"
#include"algorithm"
#include"string.h"
#include"string"
#include"cmath"
#include"queue"
#include"stack"
#include"map"
using namespace std;
const int mx=105;
const int inf=1000000;
int n,m;
int maze[mx][mx];
int MIN(int a,int b,int c)
{
    int d=a<b?a:b;
    return  c<d?c:d;
}
void DP()
{
    int i,j;
    for(j=m-1;j>=1;j--)
    {
        for(i=0;i<n;i++)
        {
            maze[i][j-1]+=MIN(maze[(i+n-1)%n][j],maze[i][j],maze[(i+1)%n][j]);
        }
    }
}
void OUTPUT()
{
    int i,j,min_length=inf,path;
    for(i=0;i<n;i++)
    {
        if(maze[i][0]<min_length)
        {
            min_length=maze[i][0];
            path=i;
        }
    }
    cout<<path+1;
    for(j=1;j<m;j++)
    {
        int r=path;
        int min_num=maze[path][j];
        if(maze[(path+n-1)%n][j]==min_num)
        {
            if((path+n-1)%n<r) r=(path+n-1)%n;
        }
        else  if(maze[(path+n-1)%n][j]<min_num)
        {
            min_num=maze[(path+n-1)%n][j];
            r=(path+n-1)%n;
        }

        if(maze[(path+1)%n][j]==min_num)
        {
            if((path+1)%n<r) r=(path+1)%n;
        }
        else if(maze[(path+1)%n][j]<min_num)
        {
            r=(path+1)%n;
            min_num=maze[(path+1)%n][j];
        }
        path=r;
        cout<<" "<<path+1;
    }
    cout<<endl<<min_length<<endl;
}
int main()
{
 //   freopen("E:\in.txt","r",stdin);
    int i,j;
    while(scanf("%d%d",&n,&m)==2)
    {
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%d",&maze[i][j]);
            }
        }
        DP();
        OUTPUT();
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/acm-jing/p/4792660.html