[Luogu P1367] ant

1 EG:  POJ Ants  　　P1007 single-plank bridge

Army ants horizontal length lcm walking pole, each having a constant velocity 1cm / s of. When reaching the end of a rod ant walking, it immediately falling. When two ants meet, they returned and began walking in the opposite direction. We know that ants original position on the pole, unfortunately, we do not know the direction of ants walking. Your task is to calculate all the ants from the shortest and the longest possible time required for the pole.

Analysis: As each ant speed equal, between ants are exactly the same, two ants met then reverse, can be seen as continue to move forward after the encounter, then all the ants from the pole shortest possible time, is each ants along the longest time period from the bar near the walking, the longest time all the ants from the pole, is the maximum time period of each ant along the rod from far away.

a = max (min (v [i] - 0 , l + 1 - v [i]), a); 
mash = max (max (v [i] - 0 , l + 1 - v [i]), mash)

 

2 EG: Luogu P1367 ants

There are many ants on an infinitely long pole, each ant has an initial position and the initial orientation (different initial positions of any two ants). Ants moving forward at a rate of one unit per second when two ants meet, they will turn around (turn around time is negligible). Are now given the initial position and the initial orientation of each ant, you calculate their position and orientation after t seconds.

 

Analysis: In this question, we will not be able to be seen as the same as each ant, but we found that the relative position of each ant order before and after the movement is constant, this can be solved.

 

for ( int I = . 1 ; I <= n-; I ++ ) { 
    CIN >> K [I] II.A >> K [I] .B; 
    R & lt [I] = K [I]; 
    R & lt [I] .order = I; 
    K [I] II.A + T * = K [I] .B; 
    R & lt [I] .bump = K [I] .bump = 0 ; 
} 
    
Sort (K + . 1 , K + . 1 + n-, CMP); // by sorting final position Sort (R & lt +. 1, n-R & lt + +. 1, CMP); // Sort the initial position 
    
for ( int I = . 1 ; I <= n-; I ++ ) { 
        K [I] .order = R & lt [I] .order; //得到初始位置的顺序 
}
    
for(int i=1;i<=n;i++){
    if(k[i].a == k[i+1].a)
        k[i].bump = 1,k[i+1].bump = 1;
}
    
sort(k+1,k+1+n,cmpp);
    
for(int i=1;i<=n;i++){
    if(k[i].bump)
        printf("%d 0\n",k[i].a);
    else printf("%d %d\n",k[i].a,k[i].b);
}