uva116 - Unidirectional TSP(简单动规)

来源：互联网发布：gta5画面优化设置编辑：程序博客网时间：2024/04/29 16:41

第二道动规题目，

这道题目卡了我很长时间的，就是因为字典序的问题我一直没想明白。

思路倒是不难，只要题目读懂了，这道题就能很容易的写出来代码。

代码如下：

#include <cstdio>#define M 15#define N 105#define INF 0x7fffffffint a[M][N], dp[M][N], pa[M][N];int m, n;void fun(){    for(int i = 0; i < m; i++)    dp[i][n-1] = a[i][n-1];    for(int y = n-2; y >= 0; y--)    {        for(int x = 0; x < m; x++)        {            int min = INF, minx = x;            if(x>0&&x<m-1)            {                if(min>dp[(x-1+m)%m][y+1]){ min = dp[(x-1+m)%m][y+1]; minx = (x-1+m)%m; }                if(min>dp[x][y+1]) { min = dp[x][y+1]; minx = x; }                if(min>dp[(x+1)%m][y+1]) { min = dp[(x+1)%m][y+1]; minx = (x+1)%m; }            }            else if(x==0)            {                if(min>dp[x][y+1]) { min = dp[x][y+1]; minx = x; }                if(min>dp[(x+1)%m][y+1]) { min = dp[(x+1)%m][y+1]; minx = (x+1)%m; }                if(min>dp[(x-1+m)%m][y+1]){ min = dp[(x-1+m)%m][y+1]; minx = (x-1+m)%m; }            }            else            {                if(min>dp[(x+1)%m][y+1]) { min = dp[(x+1)%m][y+1]; minx = (x+1)%m; }                if(min>dp[(x-1+m)%m][y+1]){ min = dp[(x-1+m)%m][y+1]; minx = (x-1+m)%m; }                if(min>dp[x][y+1]) { min = dp[x][y+1]; minx = x; }            }            dp[x][y] = min+a[x][y];            pa[x][y] = minx;        }    }}int main (){    while(~scanf("%d %d",&m,&n))    {        for(int i = 0; i < m; i++)            for(int j = 0; j < n; j++)                scanf("%d",&a[i][j]);        fun();        int min = INF, minx;        for(int i = 0; i < m; i++)        {            if(min>dp[i][0]){min = dp[i][0]; minx = i;}        }        for(int i = 0; i < n; i++)        {            if(i)   printf(" ");            printf("%d",minx+1);            minx = pa[minx][i];        }        printf("\n%d\n",min);    }    return 0;}