UVA10881:

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题目：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1822

题意：给出每个蚂蚁的初始状态，求时间T之后每只蚂蚁的状态

解法：如果忽略编号的话，可以把每只蚂蚁的掉头看成对穿而过，而最终的状态下，最左边的绝对是1号蚂蚁了

#include <stdio.h>#include <algorithm>using namespace std;const int maxn = 10005;struct NODE{    int no;    int s;    int t;//朝向，-1为左，0中，1右}start[maxn],end[maxn];int cmp(NODE x,NODE y){    return x.s < y.s;}char turn[][10] = {"L","Turning","R"};int order[maxn];int main(){    int t,cas = 1;    scanf("%d",&t);    while(t--)    {        int L,T,N,i;        scanf("%d%d%d",&L,&T,&N);        for(i = 0;i<N;i++)        {            char c;            start[i].no = i;            scanf("%d %c",&start[i].s,&c);            start[i].t = (c == 'L'?-1:1);            end[i].no = 0;            end[i].s = start[i].s + T*start[i].t;            end[i].t = start[i].t;        }        sort(start,start+N,cmp);        for(i = 0;i<N;i++)        order[start[i].no] = i;        sort(end,end+N,cmp);        for(i = 0;i<N-1;i++)        {            if(end[i].s == end[i+1].s)            end[i].t = end[i+1].t = 0;        }        printf("Case #%d:\n",cas++);        for(i = 0;i<N;i++)        {            int a = order[i];            if(end[a].s < 0 || end[a].s>L)            printf("Fell off\n");            else            printf("%d %s\n",end[a].s,turn[end[a].t+1]);        }        printf("\n");    }    return 0;}