POJ - 2698 贪心

经典面替换算法，每次选择最远的那个碟片请求进行替换。

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 10000 + 5;
queue<int>pos[maxn];
int a[105], driv[15];
int k, n;

int solve() {
	int ans = 0, cnt = 0;
	for(int i = 0; i < n; ++i) {
		int ok = 0;
		for(int j = 0; j < cnt; ++j) {
			if(driv[j] == a[i]) {
				ok = 1;
				break;
			}
		}
		if(!ok) {
			++ans;
			if(cnt < k) {
				driv[cnt++] = a[i];
			}
			else if(cnt == k) {
				int ind = 0, far = -1;
				for(int j = 0; j < k; ++j) {
					int p = pos[driv[j]].empty() ? inf : pos[driv[j]].front();
					if(p > far) {
						ind = j;
						far = p;
					}
				}
				driv[ind] = a[i];
			}
		}
		pos[a[i]].pop();
	}
	return ans;
}

int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &k, &n);
		for(int i = 0; i < n; ++i) {
			scanf("%d", &a[i]);
			pos[a[i]].push(i);
		}
		printf("%d\n", solve());
	}
	return 0;
}

如有不当之处欢迎指出！

巴特西

POJ - 2698 贪心

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