POJ 1160 Post Office

状态方程gaDistance[m][n] = min{gaDistance[m-1][t] + gaiSumDistance[t+1][n] }

gaiSumDistance[t][n]表示在t到n站间设立一个邮局所能达到的最小和；

 

 

/*Post Office Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9956 Accepted: 5341 Description There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates. Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum. You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office. Input Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000. Output The first line contains one integer S, which is the sum of all distances between each village and its nearest post office. Sample Input 10 5 1 2 3 6 7 9 11 22 44 50 Sample Output 9 Source IOI 2000 */ #include <stdio.h> #include "string.h" #define MAX_VILLAGE_NUMBER 310 #define MAX_POSTOFFICE_NUMBER 35 #define MAX_POSITION_NUM 10000 #define INFINATE_NUMBER 0x7FFFFFFF #define MIN(a,b) (((a) < (b)) ? (a) : (b)) int gaDistance[MAX_POSTOFFICE_NUMBER][MAX_VILLAGE_NUMBER]; int gaiSumDistance[MAX_VILLAGE_NUMBER][MAX_VILLAGE_NUMBER]; int PostOfficemain(void) { int iLoop; int iLoopV; int iLoopP; int iVillageNum; int iPostNum; int aiVillagePosition[MAX_VILLAGE_NUMBER]; scanf("%d %d",&iVillageNum,&iPostNum); for (iLoopV = 1; iLoopV <= iVillageNum; iLoopV++) { scanf("%d",&aiVillagePosition[iLoopV]); } for(iLoopV = 1 ; iLoopV < iVillageNum ; iLoopV++) { for(iLoop = iLoopV+1 ; iLoop <= iVillageNum ; iLoop++) { gaiSumDistance[iLoopV][iLoop] = gaiSumDistance[iLoopV][iLoop-1] + aiVillagePosition[iLoop] - aiVillagePosition[(iLoopV+iLoop)/2]; if(1 == iLoopV) { gaDistance[1][iLoop] = gaiSumDistance[iLoopV][iLoop]; } } } for (iLoopP = 2; iLoopP <= iPostNum; iLoopP++) { for (iLoopV = iLoopP+1; iLoopV <= iVillageNum-iPostNum+iLoopP ; iLoopV++) { gaDistance[iLoopP][iLoopV] = INFINATE_NUMBER; for (iLoop = iLoopP-1; iLoop <= iLoopV-1; iLoop++) { gaDistance[iLoopP][iLoopV] = MIN( gaDistance[iLoopP][iLoopV] , gaDistance[iLoopP-1][iLoop] + gaiSumDistance[iLoop+1][iLoopV] ); } } } printf("%d/n",gaDistance[iPostNum][iVillageNum]); return 0; }