bzoj3389:[usaco2004 dec]cleaning shifts arrangement on duty

Idea: can be greedy, can also shortest way.

Greedy writing: Because in the guarantee of legal premise, we choose the interval must be the right end point as far as possible, so we choose a legitimate and right end of the next interval is good (if there is no legitimate output-1 can). Time complexity O (NLOGN) (sort is nlogn, greed is O (n)).

#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm>using namespace std; #define MAXN 25005 int n,t,ans;int last[1000005];    struct node{int l,r; BOOL operator < (const node &a) Const{return l<a.l| | (L==A.L&AMP;&AMP;R&LT;A.R);}} A[MAXN];    inline int read () {int X=0;char ch=getchar (); for (;ch< ' 0 ' | |    Ch> ' 9 '; Ch=getchar ());    for (; ch>= ' 0 ' &&ch<= ' 9 '; Ch=getchar ()) x=x*10+ch-' 0 '; return x;}    int main () {N=read (), T=read ();    for (int i=1;i<=n;i++) A[i].l=read (), A[i].r=read ();    Sort (a+1,a+n+1); int cnt=0;    for (int i=1;i<=n;i++) if (A[I].L!=A[I+1].L) a[++cnt]=a[i];    N=cnt;int now=0;        for (int i=1;i<=n;i++) {int X=0;bool flag=0;        while (a[i].l<=now+1&&i<=n) X=max (X,A[I].R), i++,flag=1;        if (!flag) {ans=-1;break;}        if (X>now) now=x,ans++;    i--;    } if (now!=t) ans=-1;    printf ("%d\n", ans); return 0;}

The shortest way to do this: interval [l,r] means you can go from l-1 to R, then we will l-1 a weight of 1 to the edge to r, and then because the interval can have intersection, so I also need to i-1 a weight of 0 edge, Then run the shortest path with 0 as the starting point (starting with 0 because it is L-1-R). Time complexity O (TLOGT)

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Queue> #include <cmath>using namespace std; #define MAXN 1000005#define inf 1e9 int N,t,tot;int now[maxn],pre[ Maxn*2],son[maxn*2],val[maxn*2],dis[maxn];bool VIS[MAXN];    inline int read () {int X=0;char ch=getchar (); for (;ch< ' 0 ' | |    Ch> ' 9 '; Ch=getchar ());    for (; ch>= ' 0 ' &&ch<= ' 9 '; Ch=getchar ()) x=x*10+ch-' 0 '; return x;}    void Add (int a,int b,int c) {son[++tot]=b;    Pre[tot]=now[a];    Now[a]=tot; Val[tot]=c;}    void link (int a,int b,int c) {Add (a,b,c);} struct node{int id,val;    Node () {} node (int a,int b) {id=a,val=b;} BOOL operator < (const node &a) Const{return val>a.val;}}; Priority_queue<node> Heap;    void Dijkstra (int x) {memset (dis,127,sizeof (DIS)), dis[x]=0;    Heap.push (Node (x,0));        while (!heap.empty ()) {node x=heap.top (); Heap.pop (); int id=x.id,v=x.val;        if (Vis[id]) continue;vis[id]=1; for (iNT P=now[id];p; p=pre[p]) if (Dis[son[p]]>v+val[p]) Heap.push (node (son[p],dis[son[p]]=v+val[p]));    }} int main () {N=read (), T=read ();    for (int i=1,a,b;i<=n;i++) A=read (), b=read (), Link (a-1,b,1);    for (int i=1;i<=t;i++) Link (i,i-1,0);    Dijkstra (0);    if (DIS[T]&GT;1E9) puts ("1"), Else printf ("%d\n", dis[t]); return 0;}

bzoj3389:[usaco2004 dec]cleaning shifts scheduled on duty