Codeforces 586 - B/C
题目链接：http://codeforces.com/contest/586/problem/B
A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sausage and cheese sandwiches, but first, he needs to buy a sausage and some cheese.
The town where Laurenty lives in is not large. The houses in it are located in two rows, n houses in each row. Laurenty lives in the very last house of the second row. The only shop in town is placed in the first house of the first row.
The first and second rows are separated with the main avenue of the city. The adjacent houses of one row are separated by streets.
Each crosswalk of a street or an avenue has some traffic lights. In order to cross the street, you need to press a button on the traffic light, wait for a while for the green light and cross the street. Different traffic lights can have different waiting time.
The traffic light on the crosswalk from the j-th house of the i-th row to the (j + 1)-th house of the same row has waiting time equal to aij(1 ≤ i ≤ 2, 1 ≤ j ≤ n - 1). For the traffic light on the crossing from the j-th house of one row to the j-th house of another row the waiting time equals bj (1 ≤ j ≤ n). The city doesn't have any other crossings.
The boy wants to get to the store, buy the products and go back. The main avenue of the city is wide enough, so the boy wants to cross it exactly once on the way to the store and exactly once on the way back home. The boy would get bored if he had to walk the same way again, so he wants the way home to be different from the way to the store in at least one crossing.
Figure to the first sample.Help Laurenty determine the minimum total time he needs to wait at the crossroads.
The first line of the input contains integer n (2 ≤ n ≤ 50) — the number of houses in each row.
Each of the next two lines contains n - 1 space-separated integer — values aij (1 ≤ aij ≤ 100).
The last line contains n space-separated integers bj (1 ≤ bj ≤ 100).
Print a single integer — the least total time Laurenty needs to wait at the crossroads, given that he crosses the avenue only once both on his way to the store and on his way back home.
4
1 2 3
3 2 1
3 2 2 3
12
3
1 2
3 3
2 1 3
11
2
1
1
1 1
4
The first sample is shown on the figure above.
In the second sample, Laurenty's path can look as follows:
- Laurenty crosses the avenue, the waiting time is 3;
- Laurenty uses the second crossing in the first row, the waiting time is 2;
- Laurenty uses the first crossing in the first row, the waiting time is 1;
- Laurenty uses the first crossing in the first row, the waiting time is 1;
- Laurenty crosses the avenue, the waiting time is 1;
- Laurenty uses the second crossing in the second row, the waiting time is 3.
In the last sample Laurenty visits all the crossings, so the answer is 4.
因为走的路径形式要求非常严格，所以n幢房子就只有n种路径可供选择，直接暴力求出其中最小的和次小的相加即可。
#include<bits/stdc++.h> using namespace std; int n,a[3][55],b[55]={0}; int main() { cin>>n; a[1][1]=a[2][1]=0; for(int row=1;row<=2;row++) { for(int i=2;i<=n;i++) { cin>>a[row][i]; a[row][i]+=a[row][i-1]; } } for(int i=1;i<=n;i++) { cin>>b[i]; b[i]+=a[1][i]+(a[2][n]-a[2][i]); } sort(b+1,b+n+1); cout<<b[1]+b[2]<<endl; }
#include<bits/stdc++.h> using namespace std; int n,a[3][55],b[55]={0}; int main() { cin>>n; a[1][1]=a[2][1]=0; for(int row=1;row<=2;row++) { for(int i=2;i<=n;i++) { cin>>a[row][i]; a[row][i]+=a[row][i-1]; } } for(int i=1;i<=n;i++) { cin>>b[i]; b[i]+=a[1][i]+(a[2][n]-a[2][i]); } int min2=0x3f3f3f3f,min1=min2-1; for(int i=1;i<=n;i++) { if(b[i]<=min1)//b[i]<min1<min2 min2=min1,min1=b[i]; else if(b[i]<=min2)//min1<b[i]<min2 min2=b[i]; } cout<<min1+min2<<endl; }
PS.求最小和次小分别用了快排和遍历，惊奇的发现在CF的测试样例下用快排居然比遍历快一倍……看来O(Cn)这个常数C在某些情况下也是有着明显的作用的。
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题目链接：http://codeforces.com/contest/586/problem/C
Gennady is one of the best child dentists in Berland. Today n children got an appointment with him, they lined up in front of his office.
All children love to cry loudly at the reception at the dentist. We enumerate the children with integers from 1 to n in the order they go in the line. Every child is associated with the value of his cofidence pi. The children take turns one after another to come into the office; each time the child that is the first in the line goes to the doctor.
While Gennady treats the teeth of the i-th child, the child is crying with the volume of vi. At that the confidence of the first child in the line is reduced by the amount of vi, the second one — by value vi - 1, and so on. The children in the queue after the vi-th child almost do not hear the crying, so their confidence remains unchanged.
If at any point in time the confidence of the j-th child is less than zero, he begins to cry with the volume of dj and leaves the line, running towards the exit, without going to the doctor's office. At this the confidence of all the children after the j-th one in the line is reduced by the amount of dj.
All these events occur immediately one after the other in some order. Some cries may lead to other cries, causing a chain reaction. Once in the hallway it is quiet, the child, who is first in the line, goes into the doctor's office.
Help Gennady the Dentist to determine the numbers of kids, whose teeth he will cure. Print their numbers in the chronological order.
The first line of the input contains a positive integer n (1 ≤ n ≤ 4000) — the number of kids in the line.
Next n lines contain three integers each vi, di, pi (1 ≤ vi, di, pi ≤ 106) — the volume of the cry in the doctor's office, the volume of the cry in the hall and the confidence of the i-th child.
In the first line print number k — the number of children whose teeth Gennady will cure.
In the second line print k integers — the numbers of the children who will make it to the end of the line in the increasing order.
5
4 2 2
4 1 2
5 2 4
3 3 5
5 1 2
2
1 3
5
4 5 1
5 3 9
4 1 2
2 1 8
4 1 9
4
1 2 4 5
In the first example, Gennady first treats the teeth of the first child who will cry with volume 4. The confidences of the remaining children will get equal to - 2, 1, 3, 1, respectively. Thus, the second child also cries at the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 0, 2, 0. Then the third child will go to the office, and cry with volume 5. The other children won't bear this, and with a loud cry they will run to the exit.
In the second sample, first the first child goes into the office, he will cry with volume 4. The confidence of the remaining children will be equal to 5, - 1, 6, 8. Thus, the third child will cry with the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 5, 5, 7. After that, the second child goes to the office and cry with the volume of 5. The confidences of the remaining children will be equal to 0, 3. Then the fourth child will go into the office and cry with the volume of 2. Because of this the confidence of the fifth child will be 1, and he will go into the office last.
参考http://www.cnblogs.com/qscqesze/p/4873929.html的文章（里面说的第二个坑踩得我相当的艰辛QAQ）。
题意 一堆小孩要去看牙医，小孩进入牙医之后，就会发出叫声，使得接下来的v[i]个孩子的信心分别下降v[i],v[i]-1......1这么多 如果小孩被吓跑了，他们又会叫，使得接下来的孩子发出d[i]的叫声 然后问你一共有多少人能够看病，并且是哪些人 题解： 数据范围只有4000，那就n^2暴力就好了 有两个坑点： 1.爆int 2.得v[i]减完之后，大家再一起叫d[i]的，不是边v[i]边d[i]
#include<bits/stdc++.h> typedef long long ll; const int MAXN = 4010; using namespace std; int n; ll v[MAXN],d[MAXN],p[MAXN]; bool flag[MAXN]; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%I64d%I64d%I64d",&v[i],&d[i],&p[i]); int id[MAXN],cnt=0; for(int i=1;i<=n;i++) { if(p[i]>=0)//这个孩子看牙医了 { id[cnt++]=i;//记录答案 ll de=v[i];//他的哭声对后面孩子的影响 for(int j=i+1;j<=n;j++) { flag[j]=0; if(p[j]<0) continue; flag[j]=1; //标记他后面那些孩子中谁是依然有信心的 p[j]-=de; if(de>0) de--; //信心down…… } ll sum=0;//记录到第j个孩子之前所有直接被吓跑了的孩子对第j个孩子的影响 for(int j=i+1;j<=n;j++) { if(p[j]>=0) p[j]-=sum; if(flag[j] && p[j]<0) sum+=d[j];//如果这个孩子曾经有过信心，但是现在没了，说明他现在起正式跑路，进而对他后面的孩子产生影响 } } } printf("%d\n",cnt); for(int i=0;i<cnt;i++) printf("%s%d",(i!=0)?" ":"",id[i]); }
