9.15 校內模擬賽題解報告

阿新 • • 發佈：2021-09-15

沒有摘要的摘要

9.15 校內模擬賽題解報告

9.15 校內模擬賽題解報告

扯

\[\huge \%\%\%\%\%\ stO\ \ Zxsoul\ \ Orz\ \ \%\%\%\%\% \]

把對 zxsoul 神仙的膜拜打在最前面

福利場 + 圖論專場

T1 並查集維護子樹大小

T2 不知道考的啥的水題

T3 分層圖

大概這場考試就是這樣...

關於考試過程以及一些題外話

大概是遇到一場能做的在 zxsoul 的無意提示下喜提 AK

這次的題確實是水...

開考從 T1 開始看... 讀完題發現好像可以直接寫...

想了一下可以倒序建邊然後好像就做完了...

以為自己想得太簡單了... 就先扔掉了跑去看 T2

T2 好像可以直接倒序建圖然後遍歷一遍就行了...

是不是想的太簡單了... 不管了看 T3

T3 送了五十的最短路... 後面的分沒什麼思路

然後回頭把 T1 寫了順便把 T2 也寫了...

然後把 T3 的五十寫了...

害怕 T1 不對然後開始想 T1 的暴力怎麼寫...

發現好像要寫樹剖思索了一下還是寫了(因為太閒了)

然後開始拍...

T2 實在想不出怎麼寫暴力... 畢竟感覺自己寫的就是暴力然後就沒拍...

繼續開 T3

一直沒什麼進展... 直到十一點的時候(還剩半個小時...)

(BS 旁邊)zxsoul: 這 T3 必是個分層圖...

嗯?

分層圖... 好像可以寫...

不到二十分鐘碼完造資料卡了一下好像沒什麼問題就交了

然後? 然後就過了

~~這就 AK 了???~~

得分情況

100 + 100 + 100 = 300

傳說中的福利場

題解

T1 大魔法師

暴力比正解都難寫...

考慮每條邊在那些組合裡面會成為最大值統計這種組合的個數即可

首先考慮最大的一條邊不難發現所有經過這條邊的路徑取值均為這條邊而經過這條邊的路徑數量則為這條邊兩邊的點數的乘積統計過這條邊之後所有經過這條邊的路徑都不需要再統計了所以可以直接把這條邊刪掉然後繼續統計刪除後兩棵樹的邊

具體實現的話將邊從大到小排序依次刪除統計這條邊連線的兩點所在子樹的大小計入答案即可

但是這樣子樹的大小不太好維護所以考慮從小到大排序倒序加邊顯然這與刪邊是等價的並查集維護即可

程式碼

/*
  Source: 大魔法師
  首先有個五十的暴力 n 方列舉點對 
  
  考慮正解 
  從大到小列舉刪邊 維護子樹節點數量即可 
  
  節點數量比較難搞 考慮維護倒序加邊的過程 
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define pt putchar(' ')
#define pn putchar('\n')
#define Abs(x) ((x) < 0 ? -(x) : (x))
#define Max(x, y) ((x) > (y) ? (x) : (y))
#define Min(x, y) ((x) < (y) ? (x) : (y))
#define Swap(x, y) ((x) ^= (y) ^= (x) ^= (y))
/*----------------------------------------------------------*/
const int A = 1e4 + 7;
const int B = 2e5 + 7;
const int C = 1e6 + 7;
const int D = 1e7 + 7;
const int mod = 999999937;
const int INF = 0x3f3f3f3f;
/*----------------------------------------------------------*/
inline void File() {
	freopen("magic.in", "r", stdin);
	freopen("magic.out", "w", stdout);
}
/*----------------------------------------------------------*/
int n, siz[B], fa[B], ans;
struct node {int u, v, w;} a[B];
struct edge {int v, w, nxt;} e[B << 1];
int head[B], ecnt;
/*----------------------------------------------------------*/
inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void Print(int x) {if(x < 0) putchar('-'), x = -x; if(x > 9) Print(x / 10); putchar(x % 10 ^ 48);}
/*----------------------------------------------------------*/
void add_edge(int u, int v, int w) {e[++ecnt] = (edge){v, w, head[u]}; head[u] = ecnt;}
int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);}
bool cmp(node x, node y) {return x.w < y.w;}
void Main() {
	File();
	n = read();
	for(int i = 1; i ^ n + 1; ++i) fa[i] = i, siz[i] = 1;
	for(int i = 1, x, y, z; i ^ n; ++i)
		a[i] = (node){x = read(), y = read(), z = read()}, add_edge(x, y, z), add_edge(y, x, z);
	std::sort(a + 1, a + n, cmp);
	for(int i = 1; i ^ n; ++i)
	{
		int x = find(a[i].u), y = find(a[i].v), w = a[i].w;
		fa[x] = y; ans = (ans + siz[x] * siz[y] % mod * w % mod) % mod; siz[y] += siz[x];
	}
	Print(ans);
}
/*----------------------------------------------------------*/
signed main() {Main(); return 0;}

T2 捷徑

這場考試裡最水的一道題

建反圖倒著貪即可

容易發現每個點第一次被搜到的時候是最優的打標記複雜度為 \(O(n)\) 的

/*
  Source: 捷徑
  直接建圖 然後倒著掃一遍即可 
*/
#include<cstdio>
#include<cstring>
#define pt putchar(' ')
#define pn putchar('\n')
#define Abs(x) ((x) < 0 ? -(x) : (x))
#define Max(x, y) ((x) > (y) ? (x) : (y))
#define Min(x, y) ((x) < (y) ? (x) : (y))
#define Swap(x, y) ((x) ^= (y) ^= (x) ^= (y))
/*----------------------------------------------------------*/
const int A = 1e4 + 7;
const int B = 1e5 + 7;
const int C = 1e6 + 7;
const int D = 1e7 + 7;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
/*----------------------------------------------------------*/
inline void File() {
	freopen("contra.in", "r", stdin);
	freopen("contra.out", "w", stdout);
}
/*----------------------------------------------------------*/
int n, m, ans[B];
struct edge {int v, nxt;} e[B];
int head[B], ecnt;
bool vis[B];
/*----------------------------------------------------------*/
inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void Print(int x) {if(x < 0) putchar('-'), x = -x; if(x > 9) Print(x / 10); putchar(x % 10 ^ 48);}
/*----------------------------------------------------------*/
void add_edge(int u, int v) {e[++ecnt] = (edge){v, head[u]}; head[u] = ecnt;}
void dfs(int u, int tp) {
	ans[u] = tp; vis[u] = 1;
	for(int i = head[u], v; i; i = e[i].nxt) if(!vis[v = e[i].v]) dfs(v, tp);
}
void Main() {
	File();
	n = read(); m = read();
	for(int i = 1, x, y; i ^ m + 1; ++i) x = read(), y = read(), add_edge(y, x);
	for(int i = n; i; --i) if(!vis[i]) dfs(i, i);
	for(int i = 1; i ^ n + 1; ++i) Print(ans[i]), pt; 
}
/*----------------------------------------------------------*/
signed main() {Main(); return 0;}

T3 過路費

建分層圖跑最短路即可

程式碼

/*
  Source: 過路費
*/
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define pt putchar(' ')
#define pn putchar('\n')
#define pr std::pair <int, int> 
#define mk std::make_pair 
#define Abs(x) ((x) < 0 ? -(x) : (x))
#define Max(x, y) ((x) > (y) ? (x) : (y))
#define Min(x, y) ((x) < (y) ? (x) : (y))
#define Swap(x, y) ((x) ^= (y) ^= (x) ^= (y))
/*----------------------------------------------------------*/
const int A = 1e4 + 7;
const int B = 5e4 + 7;
const int C = 1e6 + 7;
const int D = 1e7 + 7;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
/*----------------------------------------------------------*/
inline void File() {
	freopen("pass.in", "r", stdin);
	freopen("pass.out", "w", stdout);
}
/*----------------------------------------------------------*/
int n, m, K, S, T, dis[2][A * 13], ans = INF;
struct edge {int v, w, nxt;} e[C << 2];
int head[A * 13], ecnt;
bool vis[A * 13];
std::priority_queue <pr> q;
/*----------------------------------------------------------*/
inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void Print(int x) {if(x < 0) putchar('-'), x = -x; if(x > 9) Print(x / 10); putchar(x % 10 ^ 48);}
/*----------------------------------------------------------*/
void add_edge(int u, int v, int w) {e[++ecnt] = (edge){v, w, head[u]}; head[u] = ecnt;}
void dijk(int s, bool op) {
	for(int i = 0; i ^ n + 1; ++i) dis[op][i] = INF, vis[i] = 0;
	dis[op][s] = 0; q.push(mk(0, s));
	while(!q.empty())
	{
		int u = q.top().second; q.pop(); if(vis[u]) continue; vis[u] = 1;
		for(int i = head[u], v; i; i = e[i].nxt) if(dis[op][v = e[i].v] > dis[op][u] + e[i].w)
			dis[op][v] = dis[op][u] + e[i].w, q.push(mk(-dis[op][v], v));
	}
}
void dijk2(int s, bool op) {
	dis[op][s] = 0; q.push(mk(0, s));
	while(!q.empty())
	{
		int u = q.top().second; q.pop(); if(vis[u]) continue; vis[u] = 1;
		for(int i = head[u], v; i; i = e[i].nxt) if(dis[op][v = e[i].v] > dis[op][u] + e[i].w)
			dis[op][v] = dis[op][u] + e[i].w, q.push(mk(-dis[op][v], v));
	}
}
void work1() {dijk(S, 0); Print(dis[0][T]);}
void work2() {
	dijk(S, 0); dijk(T, 1);
	for(int i = 1; i <= 2 * m; i += 2)
	{
		int u = e[i + 1].v, v = e[i].v, tmp = Min(dis[0][u] + dis[1][v], dis[0][v] + dis[1][u]);
		ans = Min(ans, tmp);
	}
	Print(ans);
}
void work3() {
	for(int i = 1; i ^ m + 1; ++i)
	{
		int x = read(), y = read(), z = read();
		add_edge(x, y, z); add_edge(y, x, z);
		for(int j = 1; j ^ K + 1; ++j)
		{
			add_edge(x + j * n, y + j * n, z);
			add_edge(y + j * n, x + j * n, z);
			
			add_edge(x + (j - 1) * n, y + j * n, 0);
			add_edge(y + (j - 1) * n, x + j * n, 0);
		}
	}
	for(int i = 0; i ^ K + 1; ++i) add_edge(T + i * n, (K + 1) * n + 1, 0);
	memset(dis, 63, sizeof dis); memset(vis, 0, sizeof vis); dijk2(S, 0);
	Print(dis[0][(K + 1) * n + 1]);
}
void Main() {
	File();
	n = read(); m = read(); K = read(); S = read(); T = read();
	work3(); return ;
	if(K != 0 && K != 1) {work3(); return ;}
	for(int i = 1, x, y, z; i ^ m + 1; ++i)
		x = read(), y = read(), z = read(), add_edge(x, y, z), add_edge(y, x, z);
	if(!K) work1();
	else if(K == 1) work2();
//	else work3();
//	work3();
}
/*----------------------------------------------------------*/
signed main() {Main(); return 0;}