HDU2871--Memory Control

Problem Description

Memory units are numbered from 1 up to N.
A sequence of memory units is called a memory block.
The memory control system we consider now has four kinds of operations:
1.  Reset Reset all memory units free.
2.  New x Allocate a memory block consisted of x continuous free memory units with the least start number
3.  Free x Release the memory block which includes unit x
4.  Get x Return the start number of the xth memory block(Note that we count the memory blocks allocated from left to right)
Where 1<=x<=N.You are request to find out the output for M operations.

Input

Input contains multiple cases.
Each test case starts with two integer N,M(1<=N,M<=50000) ,indicating that there are N units of memory and M operations.
Follow by M lines,each line contains one operation as describe above.

Output

For each “Reset” operation, output “Reset Now”.
For each “New” operation, if it’s possible to allocate a memory block,
output “New at A”,where Ais the least start number,otherwise output “Reject New”.
For each “Free” operation, if it’s possible to find a memory block occupy unit x,
output “Free from A to B”,where A and B refer to the start and end number of the memory block,otherwise output “Reject Free”.
For each “Get” operation, if it’s possible to find the xth memory blocks,
output “Get at A”,where A is its start number,otherwise output “Reject Get”.
Output one blank line after each test case.

Sample Input

6 10New 2New 5New 2New 2Free 3Get 1Get 2Get 3Free 3Reset

Sample Output

New at 1Reject NewNew at 3New at 5 Free from 3 to 4Get at 1Get at 5Reject GetReject FreeReset Now

 

其他处理都比较好做。那个FREE比较难搞。。不过我一次编译将其他功能都实现了^_^不是题解~

/*清除所有内存新开X个内存，连续左边最小的位置,这里就是HOTEL的问题（len,rlen,llen,l,r)这里其实是query释放包括X的内存块,感觉这里用递归能实现返回第X个内存快的起点,这个要怎么建树？mid?mid+1，感觉这里就是区间合并的问题*/#include <iostream>#include <cstdio>#include <cstring>#include <string>#define maxn 50008#define lson 2*id,l,mid#define rson 2*id+1,mid+1,rbool flag;inline int max(int a,int b,int c){if(a<b) a=b;if(a<c) a=c;return a;}struct ST{int l,r,len,rlen,llen;//len是指连续空余的内存int key,rkey,lkey;//key是指连续占用的内存int line,rcov,lcov;//区间个数，左右覆盖情况int set;//懒惰标记。}st[4*maxn];void buildtree(int id,int l,int r){st[id].l=l;st[id].r=r;if(l==r){st[id].len=st[id].rlen=st[id].llen=1;st[id].line=st[id].rcov=st[id].lcov=0;st[id].key=st[id].rkey=st[id].lkey=0;return;}int mid=(l+r)>>1;buildtree(lson);buildtree(rson);st[id].len=st[id].rlen=st[id].llen=r-l+1;st[id].line=st[id].rcov=st[id].lcov=0;st[id].key=st[id].rkey=st[id].lkey=0;}void PushDown(int id){if(st[id].set!=-1){st[2*id].set=st[2*id+1].set=st[id].set;if(st[id].set){st[2*id].len=st[2*id].rlen=st[2*id].llen=0;st[2*id].line=st[2*id].lcov=st[2*id].rcov=1;st[2*id].key=st[2*id].lkey=st[2*id].rkey=st[2*id].r-st[2*id].l+1;st[2*id+1].len=st[2*id+1].rlen=st[2*id+1].llen=0;st[2*id+1].line=st[2*id+1].lcov=st[2*id+1].rcov=1;st[2*id+1].key=st[2*id+1].lkey=st[2*id+1].rkey=st[2*id+1].r-st[2*id+1].l+1;}else{st[2*id].len=st[2*id].rlen=st[2*id].llen=st[2*id].r-st[2*id].l+1;st[2*id].line=st[2*id].lcov=st[2*id].rcov=0;st[2*id].key=st[2*id].rkey=st[2*id].lkey=0;st[2*id+1].len=st[2*id+1].rlen=st[2*id+1].llen=st[2*id+1].r-st[2*id+1].l+1;st[2*id+1].line=st[2*id+1].lcov=st[2*id+1].rcov=0;st[2*id+1].key=st[2*id+1].lkey=st[2*id+1].rkey=0;}st[id].set=-1;}}void PushUp(int id){st[id].len=max(st[2*id].len,st[2*id+1].len,st[2*id].rlen+st[2*id+1].llen);st[id].rlen=(st[2*id+1].rlen==st[2*id+1].r-st[2*id+1].l+1?st[2*id].rlen+st[2*id+1].len:st[2*id+1].rlen);st[id].llen=(st[2*id].llen==st[2*id].r-st[2*id].l+1?st[2*id].len+st[2*id+1].llen:st[2*id].llen);st[id].key=max(st[2*id].key,st[2*id+1].key,st[2*id].rkey+st[2*id+1].lkey);st[id].rkey=(st[2*id+1].rkey==st[2*id+1].r-st[2*id+1].l+1?st[2*id].rkey+st[2*id+1].key:st[2*id+1].rkey);st[id].lkey=(st[2*id].lkey==st[2*id].r-st[2*id].l+1?st[2*id].key+st[2*id+1].lkey:st[2*id].lkey);st[id].lcov=st[2*id].lcov;st[id].rcov=st[2*id+1].rcov;st[id].line=st[2*id].line+st[2*id+1].line-(st[2*id].rcov && st[2*id+1].lcov);}void update(int id,int l,int r,int ope)//这里用来实现区间置0和区间置1{if(st[id].l == l && st[id].r == r){st[id].set = ope;if(ope){st[id].len = st[id].llen = st[id].rlen = 0;st[id].key = st[id].lkey = st[id].rkey = r-l+1;st[id].line = st[id].rcov = st[id].lcov = 1;}else {st[id].len = st[id].llen = st[id].rlen = r-l+1;st[id].key = st[id].lkey = st[id].rkey = 0;st[id].line = st[id].rcov = st[id].lcov = 0;}return;}PushDown(id);if(st[2*id].r >= r){update(2*id,l,r,ope);PushUp(id);return;}if(st[2*id+1].l <= l){update(2*id+1,l,r,ope);PushUp(id);return;}update(2*id,l,st[2*id].r,ope);update(2*id+1,st[2*id+1].l,r,ope);PushUp(id);}int query(int id,int l,int r,int cost)//这里实现了第二个任务的主体，其实只要等下多个clear就行{if(l==r) return l;PushDown(id);if(st[2*id].len >= cost){return query(2*id,l,st[2*id].r,cost);}else if(st[2*id].rlen + st[2*id+1].llen >= cost){return st[2*id].r-st[2*id].rlen+1;}else return query(2*id+1,l,r,cost);}//接下要完成一个任务，释放包括X的内存块,用递归？void clearx(int id,int l,int r){if(st[id].l==l && st[id].r==r){if(l==r && st[id].set==0){flag=false;return;}if(id&1){if(st[id/2].rkey > r-l+1){update(1,l,r,0);clearx(id/2,r+1-st[id/2].rkey,l-1);}if(st[id/2+1].lcov){update(1,l,r,0);clearx(id/2+1,st[id/2+1].l,st[id/2+1].l+st[id/2+1].lkey-1);}}else{if(st[id/2].rkey > r-l+1){update(1,l,r,0);clearx(id/2,r+1-st[id/2].rkey,l-1);}if(st[id/2-1].rcov){update(1,l,r,0);clearx(id/2-1,st[id/2-1].r-st[id/2-1].rkey+1,st[id/2-1].r);}}update(1,l,r,0);return;}PushDown(id);if(st[2*id].r >= r){clearx(2*id,l,r);PushUp(id);}else if(st[2*id+1].l <= l){clearx(2*id+1,l,r);PushUp(id);}else {clearx(2*id,l,st[2*id].r);clearx(2*id+1,st[2*id+1].l,r);PushUp(id);}}int queryx(int id,int l,int r,int x){if(l==r) return l;PushDown(id);if(st[2*id].line >= x){return query(2*id,l,st[2*id].r,x);}else{if(st[2*id].rcov && st[2*id+1].lcov) x++;return query(2*id+1,st[2*id+1].l,r,x-st[2*id].line);}}int main(){int n,m;while(scanf("%d%d",&n,&m)==2){buildtree(1,1,n);char str[10];int pos;for(int i=1;i<=m;i++){scanf("%s%d",str,&pos);if(str[0]=='N'){if(st[1].len >= pos){int fuck=query(1,1,n,pos);printf("New at %d\n",fuck);update(1,fuck,fuck+pos-1,1);}else printf("Reject New\n");}if(str[0]=='R'){update(1,1,n,0);printf("Reset now\n");}if(str[0]=='G'){if(st[1].line >= pos)printf("Get at %d\n",queryx(1,1,n,pos));else printf("Reject Get\n");}if(str[0]=='F'){flag=true;clearx(1,pos,pos);if(!flag)printf("Reject Free\n");else{printf("hehe\n");}}}}return 0;}

已经AC：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 500080#define lson id<<1,l,mid#define rson id<<1|1,mid+1,r#define lson id<<1,l,mid#define rson id<<1|1,mid+1,r#define LL long long intLL key[maxn];int s,t;struct ST{int l,r,a,b,left,right;LL maxsum,sum,lsum,rsum;}st[maxn<<2];LL max (LL a,LL b){return a>b?a:b;}void PushUp(int id){st[id].left = st[id<<1].left;st[id].right = st[id<<1|1].right;st[id].lsum = st[id<<1].lsum;st[id].rsum = st[id<<1|1].rsum;st[id].sum = st[id<<1].sum + st[id<<1|1].sum;LL ans1 = st[id<<1].maxsum,ans2 = st[id<<1|1].maxsum;if(ans1 >= ans2){st[id].maxsum = st[id<<1].maxsum;st[id].a = st[id<<1].a;st[id].b = st[id<<1].b;}else {st[id].maxsum = st[id<<1|1].maxsum;st[id].a = st[id<<1|1].a;st[id].b = st[id<<1|1].b;}//lsum,rsum,maxsum,a,b,left,right都有可能更改//lsumif(st[id].lsum == st[id<<1].sum && st[id<<1|1].lsum > 0){st[id].lsum = st[id].lsum + st[id<<1|1].lsum;st[id].left = st[id<<1|1].left;}//rsumif(st[id].rsum == st[id<<1|1].sum && st[id<<1].rsum >= 0){st[id].rsum = st[id].rsum + st[id<<1].rsum;st[id].right = st[id<<1].right;}//maxsumif((st[id<<1].rsum + st[id<<1|1].lsum > st[id].maxsum) ||   (st[id<<1].right < st[id].a && st[id<<1].rsum + st[id<<1|1].lsum == st[id].maxsum)){st[id].maxsum = st[id<<1].rsum + st[id<<1|1].lsum;st[id].a = st[id<<1].right;st[id].b = st[id<<1|1].left;}}void buildtree(int id,int l,int r){st[id].l = l,st[id].r = r;if(l == r){st[id].sum = st[id].lsum = st[id].rsum = st[id].maxsum = key[l];st[id].left = st[id].right = st[id].a = st[id].b = l;return;}int mid = (l + r) >> 1;buildtree(lson);buildtree(rson);PushUp(id);}LL query(int id,int l,int r){if(st[id].l == l && st[id].r == r){s = st[id].a;t = st[id].b;return st[id].maxsum;}if(st[id<<1].r >= r){return query(id<<1,l,r);}else if(st[id<<1|1].l <= l){return query(id<<1|1,l,r);}else {LL ans1 = query(id<<1,l,st[id<<1].r);LL ans2 = query(id<<1|1,st[id<<1|1].l,r);if(ans1 >= ans2)query(id<<1,l,st[id<<1].r);LL ans = max(ans1,ans2);if((st[id<<1].rsum + st[id<<1|1].lsum > ans) ||    (st[id<<1].rsum + st[id<<1|1].lsum == ans && st[id<<1].right < s)){s = st[id<<1].right;t = st[id<<1|1].left;}ans = max(ans,st[id<<1].rsum + st[id<<1|1].lsum);return ans;}}int main(){//freopen("in.txt","r",stdin);int n,m,cas = 0;while(scanf("%d%d",&n,&m)==2){cas++;for(int i = 1;i <= n;i++)scanf("%lld",&key[i]);buildtree(1,1,n);printf("Case %d:\n",cas);while(m--){int u,v;scanf("%d%d",&u,&v);query(1,u,v);printf("%d %d\n",s,t);}}return 0;}