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zoj2972 Hurdles of 110m
2014-11-23 20:16:30 来源: 作者: 【 】 浏览:7
Tags:zoj2972 Hurdles 110m
#include    
#include    
#define min(a,b)  a= f1)  
                    f[i][j - f1] = min(f[i][j - f1], f[i - 1][j] + t1);  
                f[i][j] = min(f[i][j], f[i - 1][j] + t2);  
                if (j + f2 <= m)  
                    f[i][j + f2] = min(f[i][j + f2], f[i - 1][j] + t3);  
                else  
                    f[i][m] = min(f[i][m], f[i - 1][j] + t3);  
            }  
        }  
  
        int minn = INF;  
        for (i = 0; i <= m; i++)  
            if (f[n][i] < minn) minn = f[n][i];  
        printf("%d\n", minn);  
    }  
    return 0;  
}  

#include 
#include 
#define min(a,b)  a= f1)
                    f[i][j - f1] = min(f[i][j - f1], f[i - 1][j] + t1);
                f[i][j] = min(f[i][j], f[i - 1][j] + t2);
                if (j + f2 <= m)
                    f[i][j + f2] = min(f[i][j + f2], f[i - 1][j] + t3);
                else
                    f[i][m] = min(f[i][m], f[i - 1][j] + t3);
            }
        }

        int minn = INF;
        for (i = 0; i <= m; i++)
            if (f[n][i] < minn) minn = f[n][i];
        printf("%d\n", minn);
    }
    return 0;
}






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