poj 1160 Post Office(DP-简单DP)
2014-07-04 15:37
459 查看
Post Office
DescriptionThere 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 betweentwo 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 villageand 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.InputYour 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 increasingorder. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.OutputThe first line contains one integer S, which is the sum of all distances between each village and its nearest post office.Sample Input
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 15380 | Accepted: 8334 |
10 5 1 2 3 6 7 9 11 22 44 50Sample Output
9
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 10010;const int INF = 1000000000;int A[310] , sum[maxn] , sumA[maxn], dp[310][40][2];int V , P;void initial(){memset(A , 0 , sizeof A);memset(sum , 0 , sizeof sum);memset(sumA , 0 , sizeof sumA);for(int i = 0; i < 310; i++){for(int j = 0; j < 40; j++){dp[i][j][0] = INF;dp[i][j][1] = INF;}}}void readcase(){for(int i = 0; i < V; i++){scanf("%d" , &A[i]);sum[A[i]] = A[i];sumA[A[i]] = 1;}for(int i = 1; i <= A[V-1]; i++){sum[i] = sum[i]+sum[i-1];sumA[i] = sumA[i]+sumA[i-1];}}void computing(){for(int i = 0; i < V; i++){dp[i][1][1] = (i+1)*A[i]-sum[A[i]];}for(int i = 1; i < V; i++){for(int k = 1; k <= min(i+1 ,P); k++){for(int j = 0; j < i; j++){if(k > 1){dp[i][k][1] = min(dp[i][k][1] , dp[j][k-1][1]+2*sum[(A[i]+A[j])/2]-sum[A[i]]-sum[A[j]]-(sumA[(A[i]+A[j])/2]-sumA[A[j]])*A[j]+(sumA[A[i]]-sumA[(A[i]+A[j])/2])*A[i]);}dp[i][k][0] = min(dp[i][k][0] , dp[j][k][1]+sum[A[i]]-sum[A[j]]-(sumA[A[i]]-sumA[A[j]])*A[j]);}}}cout << min(dp[V-1][P][0] , dp[V-1][P][1]) << endl;}int main(){while(scanf("%d%d" , &V , &P) != EOF){initial();readcase();computing();}return 0;}
相关文章推荐
- poj 1160 Post Office(DP简单题)
- POJ 1160 Post Office (经典dp)
- POJ 1160 Post Office (二维DP)
- POJ 1160 Post Office (水DP)
- POJ 1160 Post Office(dp)
- poj 1160 Post Office & SCAU 07校赛10320 Post Office ( dp )
- POJ 1160 Post Office (DP)
- POJ 1160 Post Office(四边形不等式优化DP)
- poj 1160 Post Office(经典dp)
- [dp] poj1160 Post office
- POJ 1160 Post Office(DP)
- 【DP】 POJ 1160 Post Office
- poj 1160 Post Office (区间DP)
- poj 1160 Post Office (间隔DP)
- POJ 1160 Post Office(抽象的二维DP)
- POJ 1160 Post Office(区间DP)
- poj 1160 Post Office(DP)
- poj 1160 Post Office(dp)
- DP 【POJ1160】POST OFFICE 邮局问题
- POJ 1160 Post Office(经典DP)