Description
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The last block consists of just one line containing 0.
Output
Sample Input
5 1 2 3 4 5 5 4 1 2 3 0 6 6 5 4 3 2 1 0 0
Sample Output
Yes No Yes
分析:
通過棧來模擬,看看能否按照目標順序進棧和出棧,如果能夠模擬出目標順序的進棧與出棧,那麼模擬成功後,棧內必然爲空,如果模擬失敗,那麼棧內必然會存在元素,然後就可以根據棧是否爲空來輸出Yes或No
代碼:#include <stack>
#include <cstdio>
#include <iostream>
using namespace std;
#define MAXN (1000 + 10)
int n, target[MAXN];
int main(int argc, const char * argv[]) {
while (scanf("%d", &n) && n != 0) {
//輸入目標順序
while (scanf("%d", &target[0])) {
if (target[0] == 0) { //如果組內第一個數爲0,說明結束本組
printf("\n");
break;
} else {
for (int i = 1; i < n; i++) {
scanf("%d", &target[i]);
}
}
int i, j;
int mark = 1;
stack<int> s;
//通過棧來模擬,看看能否按照目標順序進棧和出棧,如果能夠模擬出目標順序的進棧與出棧,那麼模擬成功後,棧內必然爲空,如果模擬失敗,那麼棧內必然會存在元素,然後就可以根據棧是否爲空來輸出Yes或No
for (i = 0 ; i < n; i++) {
if (!s.empty() && s.top() == target[i]) {
s.pop();
} else {
for (j = mark ; j <= n; j++) {
s.push(j);
mark++;
if (s.top() == target[i]) {
s.pop();
break;
}
}
}
}
//如果棧爲空,模擬目標順序進棧出棧成功,輸出Yes,否則,輸出No
if (s.empty()) {
printf("Yes\n");
} else {
printf("No\n");
}
}
}
return 0;
}