Zoj 3465 The Hive (模拟题)

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4230

题目大意:给定一个峰巢结构，让我们模拟小蜜蜂产密。n只蜜蜂产26种蜜并指定在某列产蜜，一旦在某列产蜜，蜜就从顶部跑到底部。如果那列满了，就永远也装不进去了。如果上下列蜜的种类相同则产生一种candy，问最后可产多少candy？

解题思路：简单模拟题，先把蜂巢结构打表打出来，然后每接受一个输入，就用栈来模拟，看是要进栈还是退栈，最后输出有蜜的蜂巢。

测试数据：14
EA
EB
EC
ED
EE
EF
EG
EH
EI
EJ
EK
EK
EG
EL

11
EA
EA
EA
EA
EA
EA
EA
EA
EA
EA
EA

11
DA
DA
DA
DA
DA
DA
DA
DA
DA
DA
DA

11
IA
IA
IA
IA
IA
IA
IA
IA
IA
IA
IA

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int st[40][40];
int top[40],n,m,ans;
char candy[11000];
char hive[40][40];
char ori[40][40] = {
	"		",
	"         _",
    "       _/ \\_",
    "     _/ \\_/ \\_",
    "   _/ \\_/ \\_/ \\_",
    " _/ \\_/ \\_/ \\_/ \\_",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "/ \\_/ \\_/ \\_/ \\_/ \\",
    "\\_/ \\_/ \\_/ \\_/ \\_/",
    "  \\_/ \\_/ \\_/ \\_/",
    "    \\_/ \\_/ \\_/",
    "      \\_/ \\_/",
    "        \\_/",
};


int main()
{
	int i,j,k,tpcol,tp,colsum;
	
	
	
	while (scanf("%d",&n) != EOF) {
		
		ans = 0;
		memcpy(hive,ori,sizeof(ori));
		memset(top,0,sizeof(top));
		for (i = 0; i < n; ++i) {
			
			scanf("%s",candy);
			tpcol = (candy[0] - 'A' + 1) * 2 - 1;
			colsum = 11 - abs(9 - tpcol) / 2;
			tp = candy[1] - 'A';
			
			//用栈操作模拟计算糖果数
			if (top[tpcol] == 0) {
				
				top[tpcol]++;
				st[tpcol][top[tpcol]] = tp;
			}
			else if (top[tpcol] < colsum) {
				
				if (st[tpcol][top[tpcol]] == tp)
					top[tpcol]--,ans++;
				else st[tpcol][++top[tpcol]] = tp;
			}
		}


		printf("The number of candy is %d.\n",ans);
		//改变峰巢内容
		for (i = 1; i <= 17; i += 2) {//column
		
			int tpmin = 2 + abs(i-9)/2;
			int tpmax = 22 - abs(i-9)/2;
			for (k = 1,j = tpmax; k <= top[i]; j -= 2,++k)
			hive[j][i] = st[i][k] + 'A';//	printf("%d %d\n",j,i),
		}
		
		
		for (i = 1; i <= 23; ++i) 
			printf("%s\n",hive[i]);
	}
}

本文ZeroClock原创，但可以转载，因为我们是兄弟。