(ACM ICPC 2013–2014, NEERC, Northern Subregional Contest)
Flight Boarding Optimization
Input file: flight.in
Output file: flight.out
Time limit: 2 seconds
Memory limit: 256 megabytes
Peter is an executive boarding manager in Byteland airport. His job is to optimize the boarding process.
The planes in Byteland have s rows, numbered from 1 to s. Every row has six seats, labeled A to F.
There are n passengers, they form a queue and board the plane one by one. If the i-th passenger sits in
a row r i then the difficulty of boarding for him is equal to the number of passengers boarded before him
and sit in rows 1...r i −1. The total difficulty of the boarding is the sum of difficulties for all passengers.
For example, if there are ten passengers, and their seats are 6A, 4B, 2E, 5F, 2A, 3F, 1C, 10E, 8B, 5A,
in the queue order, then the difficulties of their boarding are 0, 0, 0, 2, 0, 2, 0, 7, 7, 5, and the total
difficulty is 23.
To optimize the boarding, Peter wants to divide the plane into k zones. Every zone must be a continuous
range of rows. Than the boarding process is performed in k phases. On every phase, one zone is selected
and passengers whose seats are in this zone are boarding in the order they were in the initial queue.
In the example above, if we divide the plane into two zones: rows 5–10 and rows 1–4, then during the first
phase the passengers will take seats 6A, 5F, 10E, 8B, 5A, and during the second phase the passengers
will take seats 4B, 2E, 2A, 3F, 1C, in this order. The total difficulty of the boarding will be 6.
Help Peter to find the division of the plane into k zones which minimizes the total difficulty of the
boarding, given a specific queue of passengers.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), s (1 ≤ s ≤ 1000), and k (1 ≤ k ≤ 50; k ≤ s).
The next line contains n integers r i (1 ≤ r i ≤ s).
Each row is occupied by at most 6 passengers.
Output
Output one number, the minimal possible difficulty of the boarding.
Example
flight.in flight.out
10 12 2 6
6 4 2 5 2 3 1 11 8 5
此题做的倒是一波三折，
f[i][j]=min(f[k][j-1]+w[k+1][i])
起先w[][]不会算，凭借wlm牛给的线段树思路起家，之后我调试代码，思路中断，混淆，最终还是wlm牛成功计算出了w[][]
然而在我的kn^2的朴素思想下，过了样例，草草上交，”in Queue“，CF跪了，草！
不管其他，以防万一，匆匆把代码改成了四边形优化，O(nk)，其实很像poj1160，只是为了防止超时。
靠！WA on test 5!!! 最后一刻看朴素提交竟然WA了，而四边形优化还在 in Queue，目测裸算法跪了，加优化又能如何？
吃饭，回新校区，一路上和这群acm伙伴各种hi，别是高兴啊。
挖槽！打开电脑时(笔记本已断网)，发现四边形优化 Run on test 40，惊呆了。
马上联网一看，A了，吐血啊！！~~~~~~~————————****￥￥￥%%%####
第二天又看了别人代码，也明白了正解。
题目大意：
乘客按顺序进客机，每进来一个乘客会产生不满度（difficulty），是他之前进来的且坐在他前面的乘客人数，现在可以分k个机厢，求最少difficulty
思路：
f[i][j]=min(f[k][j-1]+w[k+1][i])
w[][]求法巧妙，看正解代码即可知晓。
以下给出
线段树+四边形优化代码 O(n^2logn+nk)
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include <deque> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define INF 0x3f3f3f3f #define N 1005 using namespace std; struct Point { int r,res; }a[N]; int w[N][N],n,s,k,sum[N*N],f[N][N],ss[N][N]; bool isCalc[N]; bool cmp(Point a, Point b) { if(a.r==b.r) return a.res>b.res; return a.r<b.r; } void PushUp(int rt){ sum[rt]=sum[rt*2]+sum[rt*2+1]; } void build(int l, int r, int rt) { sum[rt]=0; if(l==r) return; int m=(l+r)/2; build(l,m,rt*2); build(m+1,r,rt*2+1); } void update(int x, int l, int r, int rt) { if(l==r) { sum[rt]=1; return; } int m=(l+r)/2; if(x<=m) update(x,l,m,rt*2); else update(x,m+1,r,rt*2+1); PushUp(rt); } int query(int x, int y, int l, int r, int rt) { if(x>y) return 0; if(x<=l&&y>=r) return sum[rt]; int m=(l+r)/2; int s=0; if(x<=m) s+=query(x,y,l,m,rt*2); if(y>m) s+=query(x,y,m+1,r,rt*2+1); return s; } int main() { // freopen("flight.in","r",stdin); // freopen("flight.out","w",stdout); scanf("%d%d%d",&n,&s,&k); for(int i=1; i<=n; i++) { scanf("%d",&a[i].r); a[i].res=i; } sort(a+1,a+n+1,cmp); int head=1,d=1; for(int i=1; i<=s; i++) { build(1,n,1); memset(isCalc,0,sizeof(isCalc)); d=i; int tmp=0; while(a[head].r<i) head++; for(int j=head; j<=n; j++) { update(a[j].res,1,n,1); while(d<a[j].r) { if(!isCalc[d]) { w[i][d]=tmp; } d++; } tmp+=query(1,a[j].res-1,1,n,1); w[i][a[j].r]=tmp; isCalc[a[j].r]=1; } while(d<=s) { if(!isCalc[d]) { w[i][d]=w[i][d-1]; } d++; } } for(int i=0; i<=s; i++) for(int j=0; j<=k; j++) f[i][j]=INF; for(int i=1; i<=s; i++) f[i][1]=w[1][i]; for(int j=2; j<=k; j++) { ss[s+1][j]=s-1; for(int i=s; i>=j; i--) { for(int l=ss[i][j-1]; l<=ss[i+1][j]; l++) if(f[i][j]>f[l][j-1]+w[l+1][i]) { f[i][j]=f[l][j-1]+w[l+1][i]; ss[i][j]=l; } } } printf("%d",f[s][k]); return 0; }
线段树+朴素代码O(n^2logn+kn^2)
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include <deque> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define INF 0x3f3f3f3f #define N 1005 using namespace std; struct Point { int r,res; }a[N]; int w[N][N],n,s,k,sum[N*N],f[N][N]; bool isCalc[N]; bool cmp(Point a, Point b) { if(a.r==b.r) return a.res>b.res; return a.r<b.r; } void PushUp(int rt){ sum[rt]=sum[rt*2]+sum[rt*2+1]; } void build(int l, int r, int rt) { sum[rt]=0; if(l==r) return; int m=(l+r)/2; build(l,m,rt*2); build(m+1,r,rt*2+1); } void update(int x, int l, int r, int rt) { if(l==r) { sum[rt]=1; return; } int m=(l+r)/2; if(x<=m) update(x,l,m,rt*2); else update(x,m+1,r,rt*2+1); PushUp(rt); } int query(int x, int y, int l, int r, int rt) { if(x>y) return 0; if(x<=l&&y>=r) return sum[rt]; int m=(l+r)/2; int s=0; if(x<=m) s+=query(x,y,l,m,rt*2); if(y>m) s+=query(x,y,m+1,r,rt*2+1); return s; } int main() { freopen("flight.in","r",stdin); freopen("flight.out","w",stdout); scanf("%d%d%d",&n,&s,&k); for(int i=1; i<=n; i++) { scanf("%d",&a[i].r); a[i].res=i; } sort(a+1,a+n+1,cmp); int head=1,d=1; for(int i=1; i<=s; i++) { build(1,n,1); memset(isCalc,0,sizeof(isCalc)); d=i; int tmp=0; while(a[head].r<i) head++; for(int j=head; j<=n; j++) { update(a[j].res,1,n,1); while(d<a[j].r) { if(!isCalc[d]) { w[i][d]=tmp; } d++; } tmp+=query(1,a[j].res-1,1,n,1); w[i][a[j].r]=tmp; isCalc[a[j].r]=1; } while(d<=s) { if(!isCalc[d]) { w[i][d]=w[i][d-1]; } d++; } } for(int i=0; i<=s; i++) for(int j=0; j<=k; j++) f[i][j]=INF; for(int i=1; i<=s; i++) f[i][1]=w[1][i]; for(int i=2; i<=s; i++) { for(int j=2; j<=min(i,k); j++) for(int l=1; l<i; l++) { if(f[i][j]>f[l][j-1]+w[l+1][i]) f[i][j]=f[l][j-1]+w[l+1][i]; } } printf("%d",f[s][k]); return 0; }
正解代码O(n^2+nk)
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include <deque> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define INF 0x3f3f3f3f #define N 1005 using namespace std; int w[N][N],n,s,k,f[N][N],ss[N][N],a[N]; int main() { // freopen("flight.in","r",stdin); // freopen("flight.out","w",stdout); scanf("%d%d%d",&n,&s,&k); for(int i=1; i<=n; i++) scanf("%d",&a[i]); for(int i=1; i<=n; i++) for(int j=i; j<=n; j++) if(a[i]<a[j]) w[a[i]][a[j]]++; //w[i][j]表示第i排对第j排的difficulty for(int j=1; j<=s; j++) { for(int i=1; i<=j; i++) w[i][j]+=w[i-1][j]; //w[i][j]表示前i排对第j排的difficulty int sum=w[j][j]; for(int i=j; i>=1; i--) w[i][j]=sum-w[i-1][j]; //w[i][j]表示第i-j排对第j排的difficlty } for(int i=1; i<=s; i++) for(int j=i; j<=s; j++) w[i][j]+=w[i][j-1]; //w[i][j]表示第i-j排对第i-j排的difficulty // for(int i=1; i<=s; i++) // for(int j=i; j<=s; j++) // printf("%d %d %d ",i,j,w[i][j]); for(int i=0; i<=s; i++) for(int j=0; j<=k; j++) f[i][j]=INF; for(int i=1; i<=s; i++) f[i][1]=w[1][i]; for(int j=2; j<=k; j++) { ss[s+1][j]=s-1; for(int i=s; i>=j; i--) { for(int l=ss[i][j-1]; l<=ss[i+1][j]; l++) if(f[i][j]>f[l][j-1]+w[l+1][i]) { f[i][j]=f[l][j-1]+w[l+1][i]; ss[i][j]=l; } } } printf("%d",f[s][k]); return 0; }