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POJ 1160 Post Office 经典DP + 四边形加速

第一次写四边形不等式的题,现在的理解就是用各种东东缩小了k的范围,从而使复杂度降低到n^2

需要满足的条件是

对于i<i‘<j<j‘ 满足 w(i,j)+w(i‘,j‘) <= w(i,j‘) + w(i‘j)

 

#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <stack>#include <climits>using namespace std;typedef long long LL;const int maxn = 305;const int maxm = 55;int f[maxn][maxm], pos[maxn], n, m, cost[maxn][maxn];int s[maxn][maxm];int main() {    while(scanf("%d%d", &n, &m) != EOF) {        for(int i = 1;i <= n;i++) {            scanf("%d", &pos[i]);        }        memset(cost,0,sizeof(cost));        for(int i = 1;i <= n;i++) {            for(int j = i;j <= n;j++) {                for(int k = i;k <= j;k++) {                    cost[i][j] += abs(pos[k] - pos[(i + j) / 2]);                }            }        }                memset(f,0x3f,sizeof(f));        int ans = INT_MAX;        for(int i = 1;i <= n;i++) {            f[i][1] = cost[1][i];            s[i][1] = 1;        }        for(int i = 2;i <= n;i++) {            int sj = min(i, m);            s[i][sj + 1] = i - 1;            for(int j = sj; j >= 2; j--) {                for(int k = s[i - 1][j]; k <= s[i][j + 1]; k++) {                    if(f[k][j - 1] + cost[k + 1][i] < f[i][j]) {                        s[i][j] = k;                         f[i][j] = f[k][j - 1] + cost[k + 1][i];                    }                }            }        }        printf("%d\n", f[n][m]);    }    return 0;}

  

POJ 1160 Post Office 经典DP + 四边形加速