UVa 10881 - Piotr's Ants - 水题

题目描述：lrj后白书第一章第五道例题

题目分析：1）从宏观上看，蚂蚁碰撞并没有影响它们的爬行； 2） 蚂蚁的相对位置是不会变的。所以只需要处理初始状态和最后的状态即可。

下面是代码：

#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

const int maxn = 10100;

struct ant
{
int id;
int dir;
int pos;
ant() {}
ant(int a,int b,int c): id(a),pos(b),dir(c) {}
} before[maxn],after[maxn];

int T;
int l,t,n;
int order[maxn];
char direction[][10] = {"L","Turning","R"};
bool cmp(ant a,ant b)
{
return a.pos < b.pos;
}
int main()
{
int cas = 0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&l,&t,&n);
for(int i = 0; i < n; i++)
{
int pos,dir;
char ch;
scanf("%d %c",&pos,&ch);
dir = (ch == 'L') ? -1 : 1;
before[i] = ant(i,pos,dir);
after[i] = ant(0,pos+dir*t,dir);
}
//for(int i = 0; i < n; i++) printf("%d ",before[i].pos); printf("\n");
//for(int i = 0; i < n; i++) printf("%d ",after[i].pos); printf("\n");
sort(before,before+n,cmp);
for(int i = 0; i < n; i++) order[before[i].id] = i;
sort(after,after+n,cmp);
for(int i = 0; i < n-1; i++)
if(after[i].pos == after[i+1].pos) after[i].dir = after[i+1].dir = 0;
printf("Case #%d:\n",++cas);
for(int i = 0; i < n; i++)
{
int id = order[i];
if(after[id].pos > l || after[id].pos < 0) printf("Fell off\n");
else printf("%d %s\n",after[id].pos,direction[after[id].dir+1]);
}
printf("\n");
}
return 0;
}