C. Heavy Intervals
1.C. Insert and Equalize2.C. Removal of Unattractive Pairs3.D. Jumping Through Segments4.E. Good Triples5.F. Shift and Reverse6.D. Yet Another Monster Fight7.A. Constructive Problems8.C. Game with Multiset9.A. Rating Increase10.B. Swap and Delete11.A. Problemsolving Log12.B. Preparing for the Contest13.C. Quests14.D. Three Activities15.E2. Game with Marbles (Hard Version)16.cf刷题有感17.A. Anonymous Informant18.A. Forked!19.B. Make Almost Equal With Mod
20.C. Heavy Intervals
21.D. Split Plus K22.A. 202323.B. Two Divisors24.C. Training Before the Olympiad25.D. Mathematical Problem26.F. Greetings27.C. Partitioning the Array28.G. Bicycles29.E. Eat the Chip30.G. Lights31.D. Array Repetition32.D. Berserk Monsters33.E. Increasing Subsequences34.D. Very Different Array35.G. Mischievous Shooter36.B. Plus-Minus Split37.B. A Balanced Problemset38.C. Did We Get Everything Covered39.D. Find the Different Ones!40.C. Grouping Increases41.D. Good Trip42.C. Physical Education Lesson43.E. Final Countdown44.D. Divisible Pairs45.G. Vlad and Trouble at MIT46.A. Brick Wall47.B. Minimize Inversions48.C. XOR-distance49.A. Moving Chips50.B. Monsters Attack!51.C. Find B52.D. Slimes53.C. Turtle Fingers: Count the Values of k54.D. Turtle Tenacity: Continual Mods55.D. Vlad and Division56.C. LR-remainders57.D. Lonely Mountain Dungeons简化题目
有一些左端点,右端点和权重,总价值是区间的长度乘上权重(区间长度=右端点-左端点),求最小总价值
开始思考
事实1.所有区间长度加起来是个定值
开始思考:能不能贪心做?答案是能。
在贪心的情况下,交换任意两个区间的端点或权重都会使总价值上升。(可以简化到只有n=2的情况考虑)
细节
在求最短区间长度的时候注意,每个区间最短长度一定是右端点减去最近没有被用过的左端点,因此栈结构呼之欲出
但只是求了每个区间的最短值,并没有求出哪个是最短区间
代码
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll c[100005]={0},dis[100005]={0};
struct unit
{
int pos;
int id;
}edge[200005];
bool cmp(unit a,unit b)
{
return a.pos<b.pos;
}
int main()
{
int t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
int len=0;
for(int i=1;i<=n;i++)
{
cin>>edge[i].pos;
edge[i].id=1;
}
for(int i=1+n;i<=2*n;i++)
{
cin>>edge[i].pos;
edge[i].id=2;
}
sort(edge+1,edge+1+2*n,cmp);
for(int i=1;i<=n;i++)cin>>c[i];
sort(c+1,c+n+1);
stack<int> q;
for(int i=1;i<=2*n;i++)
{
if(edge[i].id==2)
{
dis[++len]=edge[i].pos-q.top();
q.pop();
}
else q.push(edge[i].pos);
}
sort(dis+1,dis+n+1);
ll ans=0;
for(int i=1;i<=n;i++)ans+=dis[i]*c[n-i+1];
cout<<ans<<endl;
}
return 0;
}
合集:
codeforces
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