[Greedy] poj 2376 cleaning Shifts

Source: Internet
Author: User

Test instructions

For n segments, ask you to cover every point in [1,t], with a minimum number of segments.

Ideas:

Greedy, ascending in the x-coordinate, descending in the y-coordinate.

The first one is first placed by default.

It then continues to place the line segments based on the dependencies.

Note that [3,4] and [the] can be covered full [1,4].

Then notice that because the default is placed first, the state of the first is to be given a special sentence.

Including the first one if [1,t] Output 1.

Code:

`#include "cstdlib" #include "Cstdio" #include "CString" #include "Cmath" #include "queue" #include "algorithm" #include " iostream "#include" map "using namespace std;struct node{int x, y;}        P[26000];int CMP (node A,node b) {if (a.x==b.x) {if (A.Y&GT;B.Y) return 1;    return 0;    } if (a.x<b.x) return 1; return 0;}    int main () {int n,m;        while (scanf ("%d%d", &n,&m)!=-1) {for (int i=0;i<n;i++) scanf ("%d%d", &p[i].x,&p[i].y);        Sort (p,p+n,cmp);            if (p[0].x!=1) {puts ("-1");        Continue        } int xx=0;        int yy=p[0].y;        int Ans=1;            for (int i=1;i<n;i++) {if (P[I].Y&LT;YY) continue;            if (yy>=m) break;                if (p[i].x>yy+1) {ans=-1;            Break                } if (p[i].x>xx+1) {ans++;                Xx=yy;            YY=P[I].Y;       } else {         YY=P[I].Y;        }} if (Yy<m) ans=-1;    printf ("%d\n", ans); } return 0;}`

[Greedy] poj 2376 cleaning Shifts

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