uva11423 - Cache Simulator 树状数组

最新推荐文章于 2019-03-04 07:28:40 发布

原创最新推荐文章于 2019-03-04 07:28:40 发布 · 398 阅读

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本文深入探讨了缓存的工作原理、LRU算法及其在不同缓存大小下的表现，通过实例解析了数据引用与缓存命中率之间的关系，并介绍了如何通过输入指令（如ADDR、RANGE、STAT）来优化缓存性能。

Problem G
Cache Simulator
Input: Standard Input

Output: Standard Output

Cache is a special type of memoryproviding very small latency for accessing. The basic idea of using cache is tokeep recently used data close to processors. So, modern processors use smallcache along with large main memory. Program performance depends on the cacheperformance. If data is found in cache (Hit), execution can proceed normally.On the other hand if data is not found in cache, execution stalls until thatdata is fetched from main memory. Computer architects alwayslooks for miss rate of cache (#of misses per 100 accesses).

When a miss occurs, the newdata is kept in the cache and some existing data is removed from cache. Thereare several policies to select which one to evict. A common technique used inany computer systems is LRU (Least Recently Used). According to this scheme,the data that has not been used for longest time is evicted. Let us explain howthis scheme works –

Assume a cache with 4 entries.So, if the processor accesses data 1, 2, 3, 4, 5, 2, 3, 99, 2, 5, 4 thefollowing things will occur -

Cache	Hit/Miss	Data evicted
	Miss (1)
1	Miss (2)
1 2	Miss (3)
1 2 3	Miss (4)
1 2 3 4	Miss (5)	1
2 3 4 5	Hit (2)
2 3 4 5	Hit (3)
2 3 4 5	Miss (99)	4
2 3 5 99	Hit (2)
2 3 5 99	Hit (5)
2 3 5 99	Miss (4)	3
2 4 5 99

Here for 11 data references, wehave 7 misses. In this problem you have to find the number of misses fordifferent cache sizes. You will be given data references in several ways –

a) ADDR x : Processor will access data x

b) RANGE b y n:Processor will access all data references in the form b+y*k,where k is 0 to n-1

Your will also be givencommands STAT to print the number of misses for each cache (excluding themisses occurred before last STAT command). The above example can be abstractedin the following way –

RANGE 1 1 5

RANGE 2 1 2

ADDR 99

STAT ---------------------------------- 6

ADDR 2

RANGE 5 –1 2

STAT ----------------------------------1

Input

Your program starts with apositive integer N (1<=N<=30), which is the number of caches. The nextline contains size of each cache (in increasing order). Cache size will be atleast 2 but not exceeding 2²⁰. The next few lines will contain anyof RANGE, ADDR, or STAT. The last line will contain STAT. Total number of datareferences (all data accessed by processor) will not exceed 10⁷. Forthe sample input, total number of data references is 30 (5+2+1+1+2+10+5+4). Also,the processor uses 24-bit address, no data reference will exceed 2²⁴-1and will be non-negative. The value of b, y, n (in RANGE), and x (in ADDR) willbe consistent with all restrictions. Input is terminated by a line containingEND. The input file has around 20000 lines of inputs.

Output

For each STAT command just print thenumber of misses (as described above) in a line.

Sample Input Output for SampleInput

4 8

RANGE 1 1 5

RANGE 2 1 2

ADDR   99

STAT

ADDR   2

RANGE 5 -1 2

STAT

RANGE 0 10000 10

RANGE 0 20000 5

RANGE 0 30000 4

STAT

END

6 6

1 0

18 13

有N个cash，工作方式就是找当前地址在不在cash中，如果在就把它放到最后面，否则MISS，把最前面那个从cash中删掉，把它放到最后面。每个cash容量不一样大，RANGE操作是查询b+y*k,where k is 0 to n-1，ADDR是查询x，STAT输出每个cash MISS量。

用树状数组c[i]表示第i个查询的数。vis[x]表示x最近一次出现的位置，那么如果最近出现和这次出现中间的数小于等于cash数说明命中，否则未命中。如果再次出现就要把上次出现的删掉，add(vis[x],-1)。每次把当前出现的加上add(vis[x]=++now,1)。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXNODE 105
#define MOD 10000007
#define SIGMA_SIZE 4
typedef long long LL;
using namespace std;

const int MAXN=17000000;
const int MAXM=35;

int N,now;
int c[MAXN],vis[MAXN],cash[MAXM],ans[MAXM];

int lowbit(int x){
    return x&(-x);
}

int add(int x,int v){
    while(x<MAXN){
        c[x]+=v;
        x+=lowbit(x);
    }
}

int sum(int x){
    int ret=0;
    while(x>0){
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}

void solve(int x){
    if(vis[x]){
        int len=sum(now)-sum(vis[x]-1);
        for(int i=1;i<=N;i++){
            if(cash[i]<len) ans[i]++;
            else break;
        }
        add(vis[x],-1);
    }
    else{
        for(int i=1;i<=N;i++) ans[i]++;
    }
    add(vis[x]=++now,1);
}

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d",&N)!=EOF){
        now=0;
        memset(vis,0,sizeof(vis));
        memset(ans,0,sizeof(ans));
        memset(c,0,sizeof(c));
        char str[20];
        for(int i=1;i<=N;i++) scanf("%d",&cash[i]);
        while(scanf("%s",str),str[0]!='E'){
            int b,y,n;
            if(str[0]=='A'){
                scanf("%d",&b);
                solve(b);
            }
            else if(str[0]=='R'){
                scanf("%d%d%d",&b,&y,&n);
                for(int i=0;i<n;i++) solve(b+i*y);
            }
            else{
                for(int i=1;i<N;i++) printf("%d ",ans[i]);
                printf("%d\n",ans[N]);
                memset(ans,0,sizeof(ans));
            }
        }
    }
    return 0;
}