Difference between revisions of "ACM-ICPC 2017 Asia Xi'an"
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Xiejiadong (talk | contribs) (Created page with "== Problem A == Unsolved. == Problem B == Solved by dreamcloud. == Problem C == Unsolved. == Problem D == Unsolved. == Problem E == Solved by Xiejiadong && dreamcloud...") |
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Solved by Xiejiadong && dreamcloud && oxx1108. | Solved by Xiejiadong && dreamcloud && oxx1108. | ||
+ | |||
+ | 题意:在图上任意的添加边权为$1$,使得$\sum (a[i]-dist[i])$最小。 | ||
+ | |||
+ | 题解:建图,最小割。 | ||
+ | |||
+ | 设源点为$S$,汇点为$T$。 | ||
+ | |||
+ | 对于原图中每个点$u$,设其在原图中与$1$的距离为$dist[u]$。在新图中,$u$被拆为$u_0,u_1,$…$,u_{dist[u]}$。$ui$向$u_{i+1}$连一条容量为$(A[u]−(i+1))^2$的边,$udist[u]$向$T$连一条容量为无穷的边。如果$u$不是$1$,则$S$向$u_0$连一条无穷的边,否则连一条容量为$(A[1]−0)^2$的边。 | ||
+ | |||
+ | 对于原图中的每条无向边$(u,v)$,在新图中连若干条形如$(u_i,v_{i−1})$和$(v_i,u_{i−1})$的容量为无穷的有向边。 | ||
+ | |||
+ | 新图上的一个割对应了一个满足三角不等式的图,流量为该种方案的代价,因此在新图上跑个最大流即可。 | ||
== Problem F == | == Problem F == |
Revision as of 06:37, 23 September 2018
Problem A
Unsolved.
Problem B
Solved by dreamcloud.
Problem C
Unsolved.
Problem D
Unsolved.
Problem E
Solved by Xiejiadong && dreamcloud && oxx1108.
题意:在图上任意的添加边权为$1$,使得$\sum (a[i]-dist[i])$最小。
题解:建图,最小割。
设源点为$S$,汇点为$T$。
对于原图中每个点$u$,设其在原图中与$1$的距离为$dist[u]$。在新图中,$u$被拆为$u_0,u_1,$…$,u_{dist[u]}$。$ui$向$u_{i+1}$连一条容量为$(A[u]−(i+1))^2$的边,$udist[u]$向$T$连一条容量为无穷的边。如果$u$不是$1$,则$S$向$u_0$连一条无穷的边,否则连一条容量为$(A[1]−0)^2$的边。
对于原图中的每条无向边$(u,v)$,在新图中连若干条形如$(u_i,v_{i−1})$和$(v_i,u_{i−1})$的容量为无穷的有向边。
新图上的一个割对应了一个满足三角不等式的图,流量为该种方案的代价,因此在新图上跑个最大流即可。
Problem F
Unsolved.
Problem G
Solved by Xiejiadong.
Problem H
Solved by dreamcloud.
Problem I
Unsolved.
Problem J
Solved by oxx1108.
Problem K
Solved by dreamcloud.