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Zxr960115 is owner of a large farm. He feeds m cute cats and employs p feeders. There‘s a straight road across the farm and n hills along the road, numbered from 1 to n from left to right. The distance between hill i and (i - 1) is di meters. The feeders live in hill 1.
One day, the cats went out to play. Cat i went on a trip to hill hi, finished its trip at time ti, and then waited at hill hi for a feeder. The feeders must take all the cats. Each feeder goes straightly from hill 1 to n without waiting at a hill and takes all the waiting cats at each hill away. Feeders walk at a speed of 1 meter per unit time and are strong enough to take as many cats as they want.
For example, suppose we have two hills (d2 = 1) and one cat that finished its trip at time 3 at hill 2 (h1 = 2). Then if the feeder leaves hill 1 at time 2 or at time 3, he can take this cat, but if he leaves hill 1 at time 1 he can‘t take it. If the feeder leaves hill 1 at time 2, the cat waits him for 0 time units, if the feeder leaves hill 1 at time 3, the cat waits him for 1 time units.
Your task is to schedule the time leaving from hill 1 for each feeder so that the sum of the waiting time of all cats is minimized.
队伍训练时做到的题目，好不容易推出了dp公式并且想到了斜率优化，结果犯了SB错误导致一直错。。。
题目大致就是几只猫在一些地方，然后让人去收猫。。。饿还是看题吧。。。
直接对于某个位置的猫的 t 减去到这个位置的时间就好，问题就转换成了有一些猫每个猫都有一个值，然后给P个人分别一个值，然后每个猫的找到比他大的最近的那个人的值，然后相减，累加每个猫的，让总和最小。。。饿，表达稍微比较烂。。。
这里考虑到每个人的值一定是某只猫的值H，不然向下移动一点点可以更优。
对每个猫的值进行排序，然后从左到右dp，
dp[i][j]表示前i个猫，使用了j个人，的最小值。。。
然后递推的话 dp[i][j]=min{ dp[x][j-1]+(i-x)H[i]-S[i]+S[x] };
S表示H的前缀和。
然后转换一下变成 min{ dp[x][j-1]+S[x]-xH[i] } + iH[i]-S[i];
然后 Y[x]=dp[x][j-1] ,X[x]=x;
然后就是经典斜率DP的问题了。。。
代码如下：
// ━━━━━━神兽出没━━━━━━ // ┏┓ ┏┓ // ┏┛┻━━━━━━━┛┻┓ // ┃ ┃ // ┃ ━ ┃ // ████━████ ┃ // ┃ ┃ // ┃ ┻ ┃ // ┃ ┃ // ┗━┓ ┏━┛ // ┃ ┃ // ┃ ┃ // ┃ ┗━━━┓ // ┃ ┣┓ // ┃ ┏┛ // ┗┓┓┏━━━━━┳┓┏┛ // ┃┫┫ ┃┫┫ // ┗┻┛ ┗┻┛ // // ━━━━━━感觉萌萌哒━━━━━━ // Author : WhyWhy // Created Time : 2015年10月09日 星期五 18时45分52秒 // File Name : B.cpp #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MaxN=100005; int N,M,P; long long DP1[MaxN],DP2[MaxN]; long long *dp1,*dp2; long long H[MaxN]; long long S[MaxN]; long long X[MaxN],Y[MaxN],cou; long long d[MaxN]; bool better(int a,int b,long long H) { return (Y[a]-X[a]*H)<=(Y[b]-X[b]*H); } bool judge(long long x1,long long y1,long long x2,long long y2,long long x3,long long y3) { return (y1-y2)*(x2-x3)<=(y2-y3)*(x1-x2); } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ios::sync_with_stdio(false); cin>>N>>M>>P; d[1]=0; for(int i=2;i<=N;++i) { cin>>d[i]; d[i]+=d[i-1]; } long long a,b; for(int i=1;i<=M;++i) { cin>>a>>b; H[i]=b-d[a]; } dp1=DP1; dp2=DP2; N=M; sort(H+1,H+N+1); S[0]=0; for(int i=1;i<=N;++i) { S[i]=S[i-1]+H[i]; dp1[i]=H[i]*i-S[i]; } int p; long long TX,TY; P=min(P,N); for(int j=2;j<=P;++j) { cou=1; Y[0]=dp1[j-1]+S[j-1]; X[0]=j-1; p=0; for(int i=j;i<=N;++i) { while(p<cou-1 && better(p+1,p,H[i])) ++p; dp2[i]=Y[p]-X[p]*H[i]-S[i]+i*H[i]; TX=i; TY=dp1[i]+S[i]; while(cou-1>p && judge(TX,TY,X[cou-1],Y[cou-1],X[cou-2],Y[cou-2])) --cou; X[cou]=TX; Y[cou++]=TY; } swap(dp1,dp2); } cout<<dp1[N]<<endl; return 0; }
(中等) CF 311B Cats Transport，斜率优化DP。
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原文地址：http://www.cnblogs.com/whywhy/p/4865252.html
