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传送门冷静一下发现就是模拟费用流板子a,ba,ba,b取大也是一样做，记一个个数即可code

传送门考虑一个费用流做法S→ai(1,ai),ai→bi(1,0),bi→T(1,bi),ai→X(1,0),X→Y(k−l,0),Y→bi(1,0)S\rightarrow a_i(1,a_i),a_i\rightarrow b_i(1,0),b_i\rightarrow T(1,b_i),a_i\rightarrow X(1,0),X\rightarrow Y(k-l,0),Y\right...

传送门考虑先全部选sss，再改一些成eee考虑(i,i+1,k−ms−me,0)(i,i+1,k-ms-me,0)(i,i+1,k−ms−me,0)代表不改成eee(i,i+k,1,ei−si)(i,i+k,1,e_i-s_i)(i,i+k,1,ei​−si​)代表改成eee建一个虚点xxx像1−k1-k1−k连(inf,0)(inf,0)(inf,0)(str,x,k−ms,0)(s...

传送门把a,ba,ba,b排序后从小到大处理即可维护一下退流即可#include<bits/stdc++.h>using namespace std;#define cs const#define re register#define pb push_back#define pii pair<int,int>#define ll long long#de...

传送门将行列分别看做点如果被障碍分成几段的话就再建几个点行列向每段的点连(cap=1,cost=0),(1,1),(1,2)....(cap=1,cost=0),(1,1),(1,2)....(cap=1,cost=0),(1,1),(1,2)....的边答案即最小费用每次都跑一边过不去按询问的流量排序后增量做#include<bits/stdc++.h>using n...

传送门想来当初自己还是太naivenaivenaive了随便模拟一下建边跑个最大费用流就完了注意b=0的情况b=0的情况b=0的情况#include<bits/stdc++.h>using namespace std;const int RLEN=1<<20|1;inline char gc(){ static char ibuf[RLEN],*ib,*ob...

传送门洛谷垃圾数据写暴力，不开longlonglonglonglonglong都能过由于n≤1000n\le1000n≤1000所以能很简单的想到n2n^2n2建边的费用流洛谷数据只需要在建边的时候判一下val≤wval\le wval≤w就可以过考虑把绝对值拆开对2种情况分别以权值为下标建主席树建边即可#include<bits/stdc++.h>using na...