HDU 5289 Assignment (monotone queue)

Test instructions: to the T-foot data, and then each group of N and K, indicating the number of N, k means the maximum allowable capacity is poor, the next n number represents the ability of n individuals, the ability to poor in K within the interval has several

Analysis: Maintain a range of maximum and minimum values, so that their difference is less than k, so the use of a monotone queue

Common monotone Queue procedure:

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath>using namespace std;    const int MAXN = 1e6+5;int a[maxn];struct node{int index; int v;}    Qd[maxn];node Qx[maxn];int Main () {int t;    scanf ("%d", &t);        while (t--) {int n,k;        scanf ("%d%d", &n,&k);        for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);        } int st1,st2,ed1,ed2;        St1=st2=ed1=ed2=1;        Long Long sum=0;        int j=1;                for (int i=1;i<=n&&j<=n;i++) {if (i==1) {qd[1].index=qx[1].index=1;            QD[1].V=QX[1].V=A[1];  } else{while (st1<=ed1) {///monotonic queue Maintenance Max if (Qd[ed1].v<=a[i])  ed1--;                Smaller than a[i] and subscript smaller than I out of queue else break;               } Qd[++ed1].v=a[i];                A[i] into the queue qd[ed1].index=i;                while (ST2&LT;=ED2) {Monotonic Queue Maintenance Minimum if (qx[ed2].v>=a[i]) ed2--;                Larger than A[i] and subscript is smaller than I out of the queue else break;               } Qx[++ed2].v=a[i];                A[i] into the queue qx[ed2].index=i; while (QD[ST1].V-QX[ST2].V&GT;=K&AMP;&AMP;ST1&LT;=ED1&AMP;&AMP;ST2&LT;=ED2)//Count {if (q                    D[ST1].INDEX==J) st1++;                    if (qx[st2].index==j) st2++;                    sum+= (I-J);                j + +;            }}} while (J<=n) {sum+= (n-j+1);        j + +;    } printf ("%i64d\n", sum); }}

Two-point monotone queue procedure:

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath>using namespace std;    const int MAXN = 1e6+5;int a[maxn];struct node{int index; int v;}        Qd[maxn];node qx[maxn];int maxc (int l,int r,int D) {//two min to find the D into the queue until the while (l<=r) {int mid= (L+R)/2;        if (Qd[mid].v==d) return mid;        else if (qd[mid].v>d) l=mid+1;    else r=mid-1; } return L;}        int minc (int l,int r,int D) {//two min to find the D into the queue so far while (l<=r) {int mid= (L+R)/2;        if (Qx[mid].v==d) return mid;        else if (qx[mid].v<d) l=mid+1;    else r=mid-1; } return L;}    int main () {int t;    scanf ("%d", &t);        while (t--) {int n,k;        scanf ("%d%d", &n,&k);        for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);        } int st1,st2,ed1,ed2;        St1=st2=ed1=ed2=1;        Long Long sum=0;        int j=1; for (int i=1;i<=n&&j<=n;i++) {if (i==1) {qd[1].Index=1;                QD[1].V=A[1];                Qx[1].index=1;            QX[1].V=A[1];                         } else{Ed1=maxc (St1,ed1,a[i]);//two points to find the D into the queue so far, maintain the maximum value qd[ed1].v=a[i];                Qd[ed1].index=i; Ed2=minc (St2,ed2,a[i]);                Two points find D into the queue so far, maintain the minimum value qx[ed2].v=a[i];                Qx[ed2].index=i; while (QD[ST1].V-QX[ST2].V&GT;=K&AMP;&AMP;ST1&LT;=ED1&AMP;&AMP;ST2&LT;=ED2)//Count {if (qd[s                    T1].INDEX==J) st1++;                    if (qx[st2].index==j) st2++;                    sum+= (I-J);                j + +;            }}} while (J<=n) {sum+= (n-j+1);        j + +;    } printf ("%i64d\n", sum); }}

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HDU 5289 Assignment (monotone queue)