五一小长假

最新推荐文章于 2025-04-28 10:06:25 发布

Faullest

最新推荐文章于 2025-04-28 10:06:25 发布

阅读量369

点赞数

分类专栏：重新努力

本文链接：https://blog.youkuaiyun.com/qq_42821571/article/details/89764037

版权

重新努力专栏收录该内容

18 篇文章

订阅专栏

今天仍旧要写作业

文章目录

T1 数塔狂想曲
- 【问题描述】
- 【Coding】
T2 噪音
- 【题目描述】
- 【Coding】
T3 market
- 【Coding】
T4 value
- 【coding】

T1 数塔狂想曲

【问题描述】

FJ有M个牛棚，编号1至M，刚开始所有牛棚都是空的。
FJ有N头牛，编号1至N，这N头牛按照编号从小到大依次排队走进牛棚，每一天只有一头奶牛走进牛棚。第i头奶牛选择走进第p[i]个牛棚。
由于奶牛是群体动物，所以每当一头奶牛x进入牛棚y之后，牛棚y里的所有奶牛们都会喊一声“欢迎欢迎，热烈欢迎”，由于声音很大，所以产生噪音，产生噪音的大小等于该牛棚里所有奶牛（包括刚进去的奶牛x在内)的数量。

FJ很讨厌噪音，所以FJ决定最多可以使用K次“清空”操作，每次“清空”操作就是选择一个牛棚，把该牛棚里所有奶牛都清理出去，那些奶牛永远消失。“清空”操作只能在噪音产生后执行。

现在的问题是：FJ应该选择如何执行“清空”操作，才能使得所有奶牛进入牛棚后所产生的噪音总和最小？

【Coding】

这个可不是很难搞，但是我的源代码丢了，所以只能用后来改过的了，如下
思路上就是将一个图跑个两遍，记录每一层的最大值与次大值。如果最大值被ban就输出次大值

#include<bits/stdc++.h>
using namespace std;
const int maxxn = 1010;

int a[maxxn];
int b[maxxn];
int smaxx[maxxn];
int maxx[maxxn];
int maxx_id[maxxn];
int n;
int m;
int v[maxxn][maxxn];
int f[maxxn][maxxn];
int ff[maxxn][maxxn];

inline int Read () {
	int xx = 0;
	int ff = 1;
	char ch = getchar();
	while (ch > '9'||ch < '0') {
		if (ch == '-') ff = -1;
		ch = getchar();
	}
	while (ch >= '0'&&ch <= '9') {
		xx = (xx << 1) + (xx << 3) + (ch ^ 48);
		ch = getchar();
	}
	return xx * ff;
}

int main () {
	freopen("tower.in","r",stdin);
	freopen("tower.out","w",stdout);
	n = Read(); m = Read();
	for (int i = 1;i <= n;i++) {
		for (int j = 1;j <= i;j++) {
			v[i][j] = Read();
		}
	}
	for (int i = 1;i <= n;i++) {
		maxx[i] = -1;
		smaxx[i] = -1;
		maxx_id[i] = -1;
	}
	f[1][1] = v[1][1];
	for (int i = 2;i <= n;i++) {
		for (int j = 1;j <= i;j++) {
			f[i][j] = max(v[i][j] + max(f[i - 1][j], f[i - 1][j - 1]),f[i][j]);
		}
	}
	for (int i = 1;i <= n;i++) {
		ff[n][i] = v[n][i];
	}
	for (int i = n - 1;i >= 1;i--) {
		for (int j = i;j >= 1;j--) {
			ff[i][j] = max(ff[i][j],v[i][j] + max(ff[i + 1][j],ff[i + 1][j + 1]));
		}
	}
	for (int i = 1;i <= n;i++) {
		for (int j = 1;j <= i;j++) {
			if (maxx[i] < f[i][j] + ff[i][j] - v[i][j]) {
				maxx[i] = f[i][j] + ff[i][j] - v[i][j];
				maxx_id[i] = j;
			}
		}
		for (int j = 1;j <= i;j++) {
			if ((smaxx[i] < f[i][j] + ff[i][j] - v[i][j]) && (j != maxx_id[i])) {
				smaxx[i] = f[i][j] + ff[i][j] - v[i][j];
			}
		}
	}
	for (int i = 1;i <= m;i++) {
		int aa = Read(); int bb = Read();
		if (maxx_id[aa] == bb) printf("%d\n",smaxx[aa]);
		else printf("%d\n",maxx[aa]);
	}
	return 0;
}

T2 噪音

【题目描述】

现在的问题是：FJ应该选择如何执行“清空”操作，才能使得所有奶牛进入牛棚后所产生的噪音总和最小？

【Coding】

如下

#include<bits/stdc++.h>
using namespace std;//这个代码，用来水四十分 
typedef long long LL;
const int maxxn = 1000;

LL sum[maxxn];
LL n,m,k;
LL f[maxxn][maxxn];
LL g[maxxn][maxxn]; //这个是用来表示这个牛棚不清除所拥有的噪音 

inline LL gget(LL x) {
	return (x * (x + 1)) /2;
}
inline LL Read() {
	LL xx = 0;
	LL ff = 1;
	char ch = getchar();
	while (ch > '9' || ch < '0') {
		if (ch == '-') ff = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0') {
		xx = (xx << 1) + (xx << 3) + (ch ^ 48);
		ch = getchar();
	}
	return xx * ff;
}

int main () {
	freopen("noise.in","r",stdin);
	freopen("noise.out","w",stdout);
	n = Read(); m = Read(); k = Read();
	for (int i = 1;i <= n;i++) {
		LL a = Read();
		sum[a]++;
	}
	for (int i = 1;i <= m;i++) {
		g[i][0] = gget(sum[i]);//这里是每个牛棚的噪音
		for (int j = 1;j <= k;j++) {
			int aa = sum[i]%(j + 1); int bb = j + 1 - aa;
			g[i][j] = aa * gget(sum[i]/(j + 1) + 1) + bb * gget(sum[i]/(j + 1));
//			cout << "______________________" <<endl;
//			cout << i << ' ' <<j <<endl;	
//			cout << g[i][j] <<endl;
		}
	}
	for (int i = 1;i <= m;i++) {
		for (int j = 0;j <= k;j++) {
			if (i == 1) f[i][j] = g[i][j];
			else f[i][j] = f[i - 1][j] + g[i][0];
//			cout << "+++++++++++++++" <<endl;
//			cout << i << ' ' <<j <<endl;
//			cout << f[i][j] << endl;
			for (int ii = 0;ii <= j;ii++) {
				f[i][j] = min(f[i][j],f[i - 1][j - ii] + g[i][ii]);
			}
//			cout << f[i][j] <<endl;
		}
	}
	printf("%d\n",f[m][k]);
	return 0;
}

T3 market

题面暂无

【Coding】

题解传送门

#include<bits/stdc++.h>
using namespace std;//这个代码，用来水四十分 
typedef long long LL;//注意正序与倒序 
const LL maxxn = 90000;//注意f的含义 
const LL maxv = 300;//注意用二分
const LL maxc = 300;

LL n,m;
LL f[maxc+10][maxxn+10];
struct node {
	LL c,w,t;
};
LL tt[maxc+10];
node a[maxc+10];

bool cc(LL x,LL y) {
	return x < y;
}
bool cmp(node x,node y) {
	return x.t < y.t;
}
inline LL Read() {
	LL xx = 0;
	LL ff = 1;
	char ch = getchar();
	while (ch > '9' || ch < '0') {
		if (ch == '-') ff = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0') {
		xx = (xx << 1) + (xx << 3) + (ch ^ 48);
		ch = getchar();
	}
	return xx * ff;
}

int main () {
	freopen("market.in","r",stdin);
	freopen("market.out","w",stdout);
	n = Read(); m = Read();
	for (LL i = 1;i <= n;i++) {
		a[i].c = Read(); a[i].w = Read(); a[i].t = Read();
		tt[i] = a[i].t;
	}
	sort(tt,tt + n + 1,cc);
	sort(a,a + n + 1,cmp);
	for (LL i = 1;i <= maxc * maxv;i++) {
		f[0][i] = 999999999999ll;
	}
	for (LL i = 1;i <= n;i++) {
		for (LL j = 1;j <= 300 * 300;j++) {
			if (j >= a[i].w)	f[i][j] = min(f[i - 1][j],f[i - 1][j - a[i].w] + a[i].c);
			else f[i][j] = f[i - 1][j];//表示c
		}
	}
	for (LL i=1; i<=n; ++i)
		for (LL j=maxc*maxv-1; j>=0; --j)
			f[i][j]=min(f[i][j],f[i][j+1]);

	for (LL i = 1;i <= m;i++) {
		LL aa = Read(); LL bb = Read();
		LL ti = upper_bound(tt + 1,tt + n + 1,aa) - tt - 1;
		LL ans = upper_bound(f[ti],f[ti] + n * 300 + 1,bb) - f[ti] - 1;//取出下标
		printf("%lld\n",ans);
	}
	return 0;
}

T4 value

【coding】

#include<bits/stdc++.h>
using namespace std;//Õâ¸ö´úÂë£¬ÓÃÀ´Ë®ËÄÊ®·Ö 
typedef long long LL;
const int maxxn = 6000;

int n;
struct node {
	int w,v;
};
node lonely[maxxn];
int f[maxxn][maxxn];

inline bool cmp (node x,node y) {
	return x.v > y.v;
}
inline int Read() {
	int xx = 0;
	int ff = 1;
	char ch = getchar();
	while (ch > '9' || ch < '0') {
		if (ch == '-') ff = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0') {
		xx = (xx << 1) + (xx << 3) + (ch ^ 48);
		ch = getchar();
	}
	return xx * ff;
}

int main () {
	freopen("value.in","r",stdin);
	freopen("value.out","w",stdout);
	n = Read();
	for (int i = 1;i <= n;i++) {
		lonely[i].w = Read(); lonely[i].v = Read();
	}
	sort(lonely + 1,lonely + 1 + n,cmp);
	f[0][0] = 0;
	f[1][1] = lonely[1].w;
	for (int i = 1;i <= n;i++) {
		for (int j = 1;j <= i;j++) {
			f[i][j] = max(f[i - 1][j],f[i - 1][j - 1] + lonely[i].w - (j - 1) * (lonely[i].v));
		}
	}
	int ans = 0;
	for (int i = 1;i <= n;i++) {
		ans = max(ans,f[n][i]);
	}
	printf("%d\n",ans);
	return 0;
}