2017.8.15 校内模拟赛

8.15 解题报告

应得300，实得 275。

丢掉那25分的原因是有一个临时变量没开long long，而是int，然后运算过程爆了int，挂了25，mmp

T1 二分

T2 图论

T3 动态规划

/*

二分第k小值
枚举检验
*/
#include <algorithm>
#include <iostream>
#include <cstdio>

#define Inline __attri\
bute__( ( optimize( "-O2" ) ) )

const int BUF = 100000010;
char Buf[BUF], *buf = Buf;

Inline void read (long long &now)
{
int temp = 0;
for (now = 0; !isdigit (*buf); ++ buf)
if (*buf == '-') temp = 1;
for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
if (temp) now = -now;
}

#define Max 200010

long long a[Max], b[Max];
long long K, N, M;

Inline bool Check (long long key)
{
long long pos = M, res = 0;
for (int i = 1; i <= N; ++ i)
{
for (; pos && a[i] * b[pos] >= key; -- pos);
res += pos;
}
return res >= K;
}

#define Judge

Inline int Main ()
{

#ifdef Judge

freopen ("number.in", "r", stdin);
freopen ("number.out", "w", stdout);
fread (buf, 1, BUF, stdin);

#endif

read (N);
read (M);
read (K);

register int i;
for (i = 1; i <= N; ++ i)
read (a[i]);
for (i = 1; i <= M; ++ i)
read (b[i]);

std :: sort (a + 1, a + 1 + N);
std :: sort (b + 1, b + 1 + M);

long long l = 0, r = a
* b[M], Mid;

long long Answer;
for (; l <= r; )
{
Mid = l + r >> 1;
if (Check (Mid))
{
Answer = Mid;
r = Mid - 1;
}
else
l = Mid + 1;
}

std :: cout << Answer - 1;
fclose (stdin);
fclose (stdout);
return 0;
}
int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;};

/*

把原题中的条件转成图论条件
即把能合成的物品连边

然后跑最短路就好了
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

#define Inline __attri\
bute__( ( optimize( "-O2" ) ) )

const int BUF = 100000010;
char Buf[BUF], *buf = Buf;

Inline void read (int &now)
{
for (now = 0; !isdigit (*buf); ++ buf);
for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
}

#define Max 10020
#define Judge

int N, M;

struct Edge
{
int to, next, w;
};

class Spfa_Type
{
private :

Edge e[Max * 10 * 3];
int list[Max], C;

bool visit[Max];
int dis[Max];

public :

Inline void Insert_edge (int u, int v, int z)
{
C ++;
e[C].to = v; e[C].next = list[u];
list[u] = C; e[C].w = z;
}

Inline void Fill_Value (const int &N)
{
for (int i = 1; i <= N; ++ i)
read (dis[i]);
}

Inline int Do_Spfa ()
{
std :: queue <int> Queue;
register int i;
memset (visit, true, sizeof visit);
for (i = 1; i <= N; ++ i)
Queue.push (i);
int now;
for (; !Queue.empty (); Queue.pop ())
{
now = Queue.front ();
visit[now] = false;
for (i = list[now]; i; i = e[i].next)
if (dis[e[i].to] > dis[now] + dis[e[i].w])
{
dis[e[i].to] = dis[now] + dis[e[i].w];
if (!visit[e[i].to])
{
Queue.push (e[i].to);
visit[e[i].to] = true;
}
}
}
return dis[1];
}
};

Spfa_Type Make;

Inline int Main ()
{

#ifdef Judge

freopen ("dwarf.in", "r", stdin);
freopen ("dwarf.out", "w", stdout);
fread (buf, 1, BUF, stdin);

#endif

read (N);
read (M);

register int i;
Make.Fill_Value (N);

int x, y, z;
for (i = 1; i <= M; ++ i)
{
read (x), read (y), read (z);
Make.Insert_edge (z, x, y);
Make.Insert_edge (y, x, z);
}

printf ("%d", Make.Do_Spfa ());
fclose (stdin);
fclose (stdout);
return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}

/*
考虑把原题转化为背包问题

根据a，b，c，d数组推出背包每个物品的cost与value

然后一遍01背包得解
*/
#include <cstdio>
#include <iostream>
#include <cstring>

#define Inline __attri\
bute__( ( optimize( "-O2" ) ) )

const int BUF = 100000010;
char Buf[BUF], *buf = Buf;

Inline void read (int &now)
{
int temp = 0;
for (now = 0; !isdigit (*buf); ++ buf)
if (*buf == '-') temp = 1;
for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
if (temp) now = -now;
}

#define Max 200000
#define Judge

int dp[Max];
int a[Max], b[Max], c[Max], d[Max];

Inline int max (int a, int b)
{
return a > b ? a : b;
}

int l[Max], value[Max], cost[Max];
int Answer;

Inline int Main ()
{

#ifdef Judge

freopen ("abcd.in", "r", stdin);
freopen ("abcd.out", "w", stdout);
fread (buf, 1, BUF, stdin);

#endif

int N, M = 0;
read (N);

register int i, j;

int Count = 0;
for (i = 1; i <= N; ++ i)
read (a[i]), read (b[i]), read (c[i]), read (d[i]);

for (i = 1; i <= N; ++ i)
{
l[i] = b[i] - a[i];
M -= c[i] * a[i];
}

for (i = 1; i <= N; ++ i)
{
Answer += a[i] * d[i];
for (j = 1; j <= l[i]; j <<= 1)
{
l[i] -= j, ++ Count;
cost[Count] = c[i] * j;
value[Count] = d[i] * j;
}
if (l[i])
{
++ Count;
cost[Count] = c[i] * l[i];
value[Count] = d[i] * l[i];
}
}

memset (dp, -0x3f, sizeof dp);
dp[0] = 0;

for (i = 1; i <= Count; ++ i)
for (j = M; j >= cost[i]; -- j)
dp[j] = max (dp[j], dp[j - cost[i]] + value[i]);

printf ("%d", Answer + dp[M]);

fclose (stdin);
fclose (stdout);

return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}