2018.1.29【 AtCoder Beginner Contest 087-D 】解题报告（TLE）（STL，队列）
D - People on a Line
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
There are N people standing on the x-axis. Let the coordinate of Person i be xi. For every i, xi is an integer between 0 and 109 (inclusive). It is possible that more than one person is standing at the same coordinate.
You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (Li,Ri,Di). This means that Person Ri is to the right of Person Li by Di units of distance, that is, xRi−xLi=Di holds.
It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x1,x2,…,xN) that is consistent with the given pieces of information.
Constraints
- 1≤N≤100 000
- 0≤M≤200 000
- 1≤Li,Ri≤N (1≤i≤M)
- 0≤Di≤10 000 (1≤i≤M)
- Li≠Ri (1≤i≤M)
- If i≠j, then (Li,Ri)≠(Lj,Rj) and (Li,Ri)≠(Rj,Lj).
- Di are integers.
Input
Input is given from Standard Input in the following format:
N M L1 R1 D1 L2 R2 D2 : LM RM DM
Output
If there exists a set of values (x1,x2,…,xN) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.
Sample Input 1
3 3 1 2 1 2 3 1 1 3 2
Sample Output 1
Yes
Some possible sets of values (x1,x2,x3) are (0,1,2) and (101,102,103).
Sample Input 2
3 3 1 2 1 2 3 1 1 3 5
Sample Output 2
No
If the first two pieces of information are correct, x3−x1=2 holds, which is contradictory to the last piece of information.
Sample Input 3
4 3 2 1 1 2 3 5 3 4 2
Sample Output 3
Yes
Sample Input 4
10 3 8 7 100 7 9 100 9 8 100
Sample Output 4
No
Sample Input 5
100 0
Sample Output 5
Yes
【题目大意】
有N点在一条数轴上，坐标未X1，X2...XnM条信息，每条信息给出Li，Ri，Di，表示Xri-Xli=Di。判断这N个点能否全部满足着M条信息不冲突。
【解题思路】
1.直接利用队列。不断遍历直至队列为空输出Yes，或者出现矛盾输出No。
区分关联点信息对和非关联点信息，对于第i条信息，若Li和Ri均提到>=2次，则该信息未关联点信息，也即与其他信息共同决定是否正确。反之则为非关联点信息，举个栗子，如果Li，Ri只提到一次，那么显而易见由于点的坐标值可以重复，那么放到任意位置满足两点之间的条件即可。另若其中Li或Ri提到两次，另一者提到一次，那么只需要满足其他信息合理，再让只提到一次的点根据两点间距离调整满足即可。
关联点的建立是因为TLE了。。不过该了以后依旧，还在想办法。
另外注意每次遍历队列时，不是新一轮的第一个点对就可以重新设定原点（可能本轮后面的值有关联性导致下一轮第一个点也受影响）在开始由于没有注意到这一点一直WA，找了好久。目前解决办法时直到遍历队列一轮弹出值为0才可以重新从设点对原点，貌似有点浪费时间。。。
2.好像可以用DFS和BFS解，比赛时没来得及想，有待尝试。。
【解题代码】（TLE代码）
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=200010;
const int mmin=-200000000;
int N,M;
int X[maxn],Li[maxn],Ri[maxn],Di[maxn];
struct message
{
int l;
int r;
int d;
int circle;
};
message a[maxn];
int vis[maxn];
queue q;
int main()
{
// freopen("in.txt","r",stdin);
while(~scanf("%d%d",&N,&M))
{
memset(vis,0,sizeof(vis));
while(!q.empty()) q.pop();
for(int i=0;i=2&&vis[a[i].r]>=2)//关联点入队，
{
q.push(a[i]);
// printf("push l=%d r=%d d=%d circle=%d\n",a[i].l,a[i].r,a[i].d,a[i].circle);
}
}
int flag=0;
// for(int i=0;iN||li>N)
{
flag=1;
break;
}
else if(X[ri]!=mmin&&X[li]!=mmin)
{
count++;
// printf("Xl[%d]=%d Xr[%d]=%d X %d\n",li,X[li],ri,X[ri],X[2]);
int tempXi=X[li]+di;
if(tempXi!=X[ri])
{
flag=1;
// printf("argue l=%d r=%d d=%d circle=%d\n",now.l,now.r,now.d,now.circle);
break;
}
}
else if(X[ri]!=mmin)
{
count++;
X[li]=X[ri]-di;
// printf("set left l=%d r=%d d=%d circle=%d\n",now.l,now.r,now.d,now.circle);
}
else if(X[li]!=mmin)
{
count++;
X[ri]=X[li]+di;
// printf("set right l=%d r=%d d=%d circle=%d\n",now.l,now.r,now.d,now.circle);
}
else if(nowcircle!=ci)
{
if(!count)
{
nowcircle=ci;
count=1;
X[li]=0;
X[ri]=X[li]+di;
// printf("set zero l=%d r=%d d=%d circle=%d\n",now.l,now.r,now.d,now.circle);
}
else
{
nowcircle=ci;
count=0;
message next;
next.l=li;
next.r=ri;
next.d=di;
next.circle=ci+1;
q.push(next);
// printf("roll back l=%d r=%d d=%d circle=%d\n",next.l,next.r,next.d,next.circle);
}
}
else
{
message next;
next.l=li;
next.r=ri;
next.d=di;
next.circle=ci+1;
q.push(next);
// printf("roll back l=%d r=%d d=%d circle=%d\n",next.l,next.r,next.d,next.circle);
}
}
if(flag) printf("No\n");
else printf("Yes\n");
}
}
【收获与反思】
哎，速度不够快，前三题是精准了些，但是太慢了啊。出来1题400名，2题300名，3题200名，这一次第四题应该比上一次简单些，但是还是没做出来。学到的东西需要融会贯通。。