标签：several png city 交换 stash war mini ase get
昨天千里万里跑回学校打那个网络赛，然并卵，没有拿到现场赛资格。
今天花了一下午一晚上时间学习树状数组，lowbit函数取x&（-x）即可得x的二进制从右往左第一个1的位置，add函数用来修改一个值以及所有包含这个元素节点的值，quiry查询是为了获得a[1]+a[2]+??????+a[x]的值，那么这个题的quiry就是找的以前的最大值加上当前的值。
YJJ‘s Salesman
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1126 Accepted Submission(s): 389
Problem Description
YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109). YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109), he will only forward to (x+1,y), (x,y+1) or (x+1,y+1).
On the rectangle map from (0,0) to (109,109), there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109), only the one who was from (xk?1,yk?1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.
Input
The first line of the input contains an integer T (1≤T≤10),which is the number of test cases.
In each case, the first line of the input contains an integer N (1≤N≤105).The following N lines, the k-th line contains 3 integers, xk,yk,vk (0≤vk≤103), which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.
Output
The maximum of dollars YJJ can get.
Sample Input
1
3
1 1 1
1 2 2
3 3 1
Sample Output
3
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define lowbit(x) x&(-x) const int mod = 1e9+7; const int mx = 1e5+5; typedef long long ll; int x[mx]; int n; ll sum[mx]; void add(int p,ll x){ while(p<=n){ sum[p] = max(sum[p],x); p += lowbit(p); } } ll query(int p){ ll ans = 0; while(p){ ans = max(ans,sum[p]); p -= lowbit(p); } return ans; } struct node{ int x,y; ll w; bool operator<(const node &a)const{ if(y != a.y) return y < a.y; return x > a.x; } }a[mx]; int main(){ int t; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i = 1; i <= n; i++){ scanf("%d%d%lld",&a[i].x,&a[i].y,&a[i].w); x[i] = a[i].x; sum[i] = 0; } sort(a+1,a+n+1); sort(x+1,x+n+1); ll res = 0; for(int i = 1; i <= n; i++){ int k = lower_bound(x+1,x+n+1,a[i].x)-x; ll ans = a[i].w+query(k-1); add(k,ans); res = max(ans,res); } printf("%lld\n",res); } return 0; }
还有一个题是pzr在做，但是他没过。。。
emmm这个题我们考虑一个新的物品跟小根堆的堆顶比较，如果新物品比堆顶还小，说明交易亏本，直接丢入堆中，否则就交易，如果堆顶的元素是已经跟之前交易过的，那么就相当于当前物品和之前那个交易，然后堆顶就变成没有交易了，继续插入堆中，交换次数不变，如果堆顶是没有交易过的，那么就交易，交易次数+2.我们希望交换次数尽可能的小，那么价值相同是，已经交换过得就有限辣
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1319 Accepted Submission(s): 434
#include <iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<queue> #include<algorithm> using namespace std; struct node { int num; int flag; node() { }; node(long long a,long long b) { num=a; flag=b; } }; bool operator <(node a,node b) { if(a.num==b.num) return a.flag>b.flag; return a.num>b.num; } int main() { int t,temp,i; int n; cin>>t; while(t--) { long long cnt=0,sum=0; cin>>n; priority_queue<node>que; for(i=1;i<=n;i++) { cin>>temp; if(que.empty()) { que.push(node(temp,1)); continue; } node ans=que.top(); if(temp>ans.num) { sum+=temp-ans.num; if(ans.flag==1) cnt+=2; que.pop(); que.push(node(temp,0)); } que.push(node(temp,1)); } cout<<sum<<" "<<cnt<<endl; while(!que.empty()) { que.pop(); } } return 0; }
这个题，，，一直超时最后他们跟我说根据费马小定理n>2的情况根本不存在，哭唧唧。
然后还有个什么勾股定理也可以简单判，那个定理长这样
所以说这道题其实是一道非常简单的题辣！
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1142 Accepted Submission(s): 313
Special Judge
#include <cstdio> #include <iostream> #include <cmath> #include <map> using namespace std; int main() { long long n; int a; int t; scanf("%d",&t); while(t--) { scanf("%lld %d",&n,&a); if(n>2) printf("-1 -1\n"); else if(n==0) printf("-1 -1\n"); else if(n==1) printf("1 %d\n",a+1); else { if(a==1||a==2) printf("-1 -1\n"); else if(a%2==1){ int sum=a*a; printf("%d %d\n",sum/2,sum/2+1); } else{ int sum=a*a/2; printf("%d %d\n",sum/2-1,sum/2+1); } } } return 0; }
呜呜呜，我好菜啊。
标签：several png city 交换 stash war mini ase get
原文地址：https://www.cnblogs.com/kepa/p/9539234.html
