POJ 1065:wooden Sticks

Greed is a soft rib ...

Do a good brush on the problem.

Or we're going to blow up tomorrow.

Let's all be greedy today.

Well!

Or the sort of a struct. (this thing is really useful!) ）

Sort lengths from small to large

If the length is equal then the weight is small in front

Select the first node is not an initial node

Then scan the side array

See those nodes can be processed without processing time

Mark it well.

And then in sequential scan side array

Find the first node without a tag

Do the same scan again.

Loop in turn

See how many times you finish

That is the time of processing

So this sort really is the essence!

I can only say that the knees are rotten ...

L and W mean the opposite, W is the length, L is the weight. #include <iostream> #include <algorithm> #include <stdio.h>using namespace std;struct wood{int w;int l ;} A[5001];bool CMP (wood A,wood b) {//if (A.W = = B.W)//return a.l < b.l;//same length, small weight in front//else//    return A.W < b.w;
   
    return A.w==b.w?a.l < B.L:A.W < B.W;} the int main () {int d[5001];//is used to record the starting point int t,n,ans,i,j;scanf ("%d", &t) of the different sequences, while (t--) {scanf ("%d", &n), and for (i=0;i <n;i++) scanf ("%d%d", &A[I].W,&A[I].L), sort (a,a+n,cmp), or//length first in ascending order, where the weight conflicts will be rescheduled ans = 1;d[0] = a[0].l;/ /first set the first to start for (i=1;i<n;i++) {for (j=0;j<ans;j++) {if (a[i].l >= d[j])//Can be in the back, update {d[j] = A[I].L;   Break;}} if (J==ans)//All starting point does not work, then re-arrange from here a starting point {if (A[i].l < d[j-1]) {d[j] = A[I].L;    Ans + +;}}} printf ("%d\n", ans);}}
   

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POJ 1065:wooden Sticks