POJ 1837

DP问题。

1.阶段:每放置一个weight为一个阶段；
2.状态:每个阶段的“力矩”和作为状态（这里“力矩”为weight*position）；
3.状态转移:每个阶段有若干种可能状态，求出下一阶段的所有可能状态并计算到达每个状态的次数；
4.目标:最后一个阶段的时，到达状态“0”的次数；

 

/*BalanceTime Limit: 1000MS Memory Limit: 30000K Total Submissions: 4526 Accepted: 2664 DescriptionGigel has a strange "balance" and he wants to poise it. Actually, the device is different from any other ordinary balance. It orders two arms of negligible weight and each arm's length is 15. Some hooks are attached to these arms and Gigel wants to hang up some weights from his collection of G weights (1 <= G <= 20) knowing that these weights have distinct values in the range 1..25. Gigel may droop any weight of any hook but he is forced to use all the weights. Finally, Gigel managed to balance the device using the experience he gained at the National Olympiad in Informatics. Now he would like to know in how many ways the device can be balanced. Knowing the repartition of the hooks and the set of the weights write a program that calculates the number of possibilities to balance the device. It is guaranteed that will exist at least one solution for each test case at the evaluation. InputThe input has the following structure: ? the first line contains the number C (2 <= C <= 20) and the number G (2 <= G <= 20); ? the next line contains C integer numbers (these numbers are also distinct and sorted in ascending order) in the range -15..15 representing the repartition of the hooks; each number represents the position relative to the center of the balance on the X axis (when no weights are attached the device is balanced and lined up to the X axis; the absolute value of the distances represents the distance between the hook and the balance center and the sign of the numbers determines the arm of the balance to which the hook is attached: '-' for the left arm and '+' for the right arm); ? on the next line there are G natural, distinct and sorted in ascending order numbers in the range 1..25 representing the weights' values. OutputThe output contains the number M representing the number of possibilities to poise the balance. Sample Input2 4-2 3 3 4 5 8Sample Output2*/#include <stdio.h>#include <string.h>#define MAX_STAGE_NUMBER 20#define MAX_STATE_NUMBER 7501 /*15*25*20*/#define STATE_ZERO 3251int giStateFrequence[MAX_STAGE_NUMBER][MAX_STATE_NUMBER] = {0};int Balancemain(void){int iPositionNum;int iWeightNum;int aiWeight[20];int aiPosition[20];int iSumWeight = 0;int iMaxPosition;int iState;int iMaxState;int iMinState;int iLoopWeight;int iLoopPosition;int iLoopState;scanf("%d %d",&iPositionNum,&iWeightNum);for (iLoopPosition = 0; iLoopPosition < iPositionNum; iLoopPosition++){scanf("%d",&aiPosition[iLoopPosition]);}for (iLoopWeight = 0; iLoopWeight < iWeightNum; iLoopWeight++){scanf("%d",&aiWeight[iLoopWeight]);iSumWeight += aiWeight[iLoopWeight];}/************************************************************************//* 用于剪枝 *//************************************************************************/iMaxPosition = (-aiPosition[0])>aiPosition[iPositionNum-1]?(-aiPosition[0]):aiPosition[iPositionNum-1];iMaxState = STATE_ZERO + iSumWeight*iMaxPosition/2 + 1;iMinState = STATE_ZERO - iSumWeight*iMaxPosition/2 - 1;if (MAX_STATE_NUMBER <= iMaxState){iMaxState = MAX_STATE_NUMBER-1;}if (0 > iMinState){iMinState = 0;}/*first weight*/for (iLoopPosition = 0; iLoopPosition < iPositionNum; iLoopPosition++){giStateFrequence[0][STATE_ZERO+aiPosition[iLoopPosition]*aiWeight[0]]++;}/*other weight*/for (iLoopWeight = 1; iLoopWeight < iWeightNum; iLoopWeight++){for (iLoopPosition = 0; iLoopPosition < iPositionNum; iLoopPosition++){for (iLoopState = iMinState; iLoopState <= iMaxState; iLoopState++){/*寻找上一阶段的状态*/if (0 != giStateFrequence[iLoopWeight-1][iLoopState]){iState = iLoopState + aiPosition[iLoopPosition]*aiWeight[iLoopWeight];/*计算本阶段状态*/if (iMinState <= iState && iMaxState >= iState ){/*有效状态*/giStateFrequence[iLoopWeight][iState] += giStateFrequence[iLoopWeight-1][iLoopState];}}}}}printf("%d/n",giStateFrequence[iWeightNum-1][STATE_ZERO]);return 0;}