uva10881解题报告

Problem D
Piotr's Ants
Time Limit: 2 seconds

"One thing is for certain: there is no stopping them;
the ants will soon be here. And I, for one, welcome our
new insect overlords."Kent Brockman

Piotr likes playing with ants. He has n of them on a horizontalpoleL cm long. Each ant is facing either left or right and walksat a constant speed of 1 cm/s. When two ants bump into each other, theyboth turn around (instantaneously) and start walking in opposite directions.Piotr knows where each of the ants starts and which direction it is facingand wants to calculate where the ants will end upT seconds from now.

Input
The first line of input gives the number of cases, N. Ntest cases follow. Each one starts with a line containing 3 integers:L ,T and n (0 <= n <= 10000).The nextn lines give the locations of the n ants (measuredin cm from the left end of the pole) and the direction they are facing(L or R).

Output
For each test case, output one line containing "Case #x:"followed byn lines describing the locations and directions of then ants in the same format and order as in the input. If two or moreants are at the same location, print "Turning" instead of "L" or "R" fortheir direction. If an ant falls off the pole before T seconds,print "Fell off" for that ant. Print an empty line after each test case.

Sample InputSample Output

210 1 41 R5 R3 L10 R10 2 34 R5 L8 R

Case #1:2 Turning6 R2 TurningFell offCase #2:3 L6 R10 R

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Problemsetter: Igor Naverniouk
Alternate solutions: Frank Pok Man Chu and Yury Kholondyrev

 

 

把蚂蚁的碰撞看成“对穿”，时间t后的所有蚂蚁位置即可确定，蚂蚁的相对顺序不变，就可以找到每个位置所对应的蚂蚁。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int N;struct ant{    int pos;    int dir;    int id;    bool operator<(const ant&t)const{        return pos<t.pos;    }}before[10010],after[10010];int ps[10010];const char* str[]={"L","R","Turning","Fell off"};int main(){    int i,j,k,m,n,l,t;    int kase;    char c;    scanf("%d",&N);    for(kase=1;kase<=N;kase++){            printf("Case #%d:\n",kase);        scanf("%d%d%d",&l,&t,&n);        for(i=0;i<n;i++){            scanf("%d %c",&j,&c);            if(c=='R'){                after[i]=(ant){j+t,1,i};            }            else {                after[i]=(ant){j-t,0,i};            }            before[i]=(ant){j,0,i};                    }        sort(before,before+n);        sort(after,after+n);      //  if(after[0].pos<0||after[0].pos>l)after[0].dir=3;        for(i=1;i<n;i++){            if(after[i].pos==after[i-1].pos){                after[i].dir=after[i-1].dir=2;            }//            if(after[i].pos<0||after[i].pos>l)after[i].dir=3;        }        for(i=0;i<n;i++){            ps[before[i].id]=i;        }        for(i=0;i<n;i++){            if(after[ps[i]].pos<0||after[ps[i]].pos>l)printf("%s\n",str[3]);            else            printf("%d %s\n",after[ps[i]].pos,str[after[ps[i]].dir]);        }        printf("\n");    }    return 0;}