HDU - 3410 Passing the Message 单调递减栈
Passing the Message
Because all kids have different height, Teacher Liu set some message passing rules as below:
1.She tells the message to the tallest kid.
2.Every kid who gets the message must retell the message to his “left messenger” and “right messenger”.
3.A kid’s “left messenger” is the kid’s tallest “left follower”.
4.A kid’s “left follower” is another kid who is on his left, shorter than him, and can be seen by him. Of course, a kid may have more than one “left follower”.
5.When a kid looks left, he can only see as far as the nearest kid who is taller than him.
The definition of “right messenger” is similar to the definition of “left messenger” except all words “left” should be replaced by words “right”.
For example, suppose the height of all kids in the row is 4, 1, 6, 3, 5, 2 (in left to right order). In this situation , teacher Liu tells the message to the 3rd kid, then the 3rd kid passes the message to the 1st kid who is his “left messenger” and the 5th kid who is his “right messenger”, and then the 1st kid tells the 2nd kid as well as the 5th kid tells the 4th kid and the 6th kid.
Your task is just to figure out the message passing route.
InputThe first line contains an integer T indicating the number of test cases, and then T test cases follows.
Each test case consists of two lines. The first line is an integer N (0< N <= 50000) which represents the number of kids. The second line lists the height of all kids, in left to right order. It is guaranteed that every kid’s height is unique and less than 2^31 – 1 .OutputFor each test case, print “Case t:” at first ( t is the case No. starting from 1 ). Then print N lines. The ith line contains two integers which indicate the position of the ith (i starts form 1 ) kid’s “left messenger” and “right messenger”. If a kid has no “left messenger” or “right messenger”, print ‘0’ instead. (The position of the leftmost kid is 1, and the position of the rightmost kid is N)Sample Input
2 5 5 2 4 3 1 5 2 1 4 3 5
Sample Output
Case 1: 0 3 0 0 2 4 0 5 0 0 Case 2: 0 2 0 0 1 4 0 0 3 0 与LC的课后辅导类似,这道用到了单调递减栈,分别找左右两端点,c记录最后一个出栈元素,递增栈退栈顶变小,递减栈退栈顶变大,第一个恰好比当前值大的前一个即为信使。
#include<stdio.h> #include<string.h> #include<stack> using namespace std; int main() {int t,tt,n,c,f,i;int a[50005],l[50005],r[50005];stack<int> s;scanf("%d",&t);tt=t;while(t--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}c=1;memset(l,0,sizeof(l));memset(r,0,sizeof(r));for(i=1;i<=n;i++){f=0;while(s.size()&&a[s.top()]<=a[i]){f=1;c=s.top();s.pop();}l[i]=f==0?0:c;s.push(i);}while(s.size()){s.pop();}c=n;for(i=n;i>=1;i--){f=0;while(s.size()&&a[s.top()]<=a[i]){f=1;c=s.top();s.pop();}r[i]=f==0?0:c;s.push(i);}while(s.size()){s.pop();}printf("Case %d:\n",tt-t);for(i=1;i<=n;i++){printf("%d %d\n",l[i],r[i]);}}return 0; }
转载于:https://www.cnblogs.com/yzm10/p/7222587.html
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