This is a standard half-wave voltage doubler circuit. The diodes and capacitors are drawn correctly. The purpose of the resistor is to limit the charging current of the two capacitors. Each diode conducts alternately on the positive and then the negative half-cycles of the power line, charging the capacitors to the peak of the line voltage or approximately 170 V DC across each capacitor. The capacitors are connected in series, so the total voltage is 170 + 170 = 340 V DC. Since you plan to use this with a xenon flash system for still (single frame) photography, the time to charge the capacitors is not very important. You could increase the resistor value to decrease the initial charging current and the power dissipated in the resistor.
If this results in too long a time to re-charge the capacitors between flashes, decrease the value and perhaps increase the wattage rating of the resistor. Most commonly, a resistor is not needed at all, but that means a large "surge" current will occur through the 1N4007 diodes which are rated for 30 A for a one-cycle surge, so assuming the capacitors will fully charge in one half-cycle of the AC line voltage, the current-limiting resistor could be as small as 170 / 30 = 6 ohms, dissipating an instantaneous power of about 5400 watts for a duration of less than 8 ms at 60 Hz line frequency... about 43 J of energy. Note this is worst case. An uncharged capacitor has zero apparent resistance, so the instantaneous current can be as much as the peak AC line voltage (170 V) divided by the value of the current-limiting resistor if power is applied at the peak of the AC line cycle.
Unfortunately, connecting this circuit directly to the mains power means there is virtually no way to limit the surge current without using a resistor. It's possible that the resistor should have been specified at 1000 ohms at 1 watt. This would limit the current to 170 /1000 = 0.17 A and an instantaneous power dissipation of about 29 watts during the half-cycle or so required to charge the capacitor... less than a quarter joule of energy to dissipate.
Once the capacitors are charged, no further current should flow through the current-limiting resistor, so power dissipation will be nil until the flash-tube discharges the capacitors and the power line voltage begins to re-charge them again. If you make the resistor too large, it will require more than one half-cycle to charge the capacitor, but the peak current always occurs first and decreases on subsequent half-cycles. Try using a 1 kΩ, 1 W resistor.
This circuit is dangerous if operated directly from mains power. You should use a 1:1 isolation transformer to provide line voltage to the diodes. If you do that you won't need the current-limiting resistor.