Contest Hunter Round #70 - 连续两大交易事件杯省选模拟赛

orz lydrainbowcat

[Problem A]「艦これ市」70万幕后交易事件

排序机器=-=。重要的是相同的处理。

我们可以从小到大添加数字，然后维护一个位置的序列。每一种相等的数字都在一块。如果我们要添加一个新的数字，要把位置>它的数字全部弹出，而且要把小于它的数字(在队头)全部弹出，这样才能保证正确性和最优性。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

#define maxn 1000010

int n, m, ans;

int a[maxn];

vector<int> b[maxn];

int q[maxn];

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++){
		scanf("%d", &a[i]);
		b[a[i]].push_back(i);
		m = max(m, a[i]);
	}
	int l = 1, r = 0;
	for(int i = 1; i <= m; i ++){
		int N = b[i].size();
		for(int j = N - 1; j >= 0; j --){
			int k = b[i][j];
			while(l <= r && q[r] > k){
			    while(l <= r && a[q[l]] < a[q[r]])
					l ++;
				r --;
   			}
			ans = max(ans, r - l + 1 + N - j);
		}
		for(int j = 0; j < b[i].size(); j ++)
		    q[++ r] = b[i][j];
	}
	printf("%d\n", n - ans);
	return 0;
}

[Problem B]「NOIP市」神秘PY交易事件（前篇）　　

题目意思好长。。

而且两个人同能力要相等的工资？？

暴力都写炸了。。QAQ。。

bruteforce:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 100010
using namespace std;

struct Edge{
	int to, next, dis;
	Edge(int v = 0, int nxt = 0, int d = 0){
		to = v, next = nxt, dis = d;
	}
}edge[maxn << 2];

int h[maxn], cnt, n;
void add(int u, int v, int d){
	edge[++ cnt] = Edge(v, h[u], d); h[u] = cnt;
}

int dis[maxn];
queue<int>Q;
bool vis[maxn];
void Spfa(){
	int ans = 0;
	for(int i = 1; i <= n; i ++)
	    add(0, i, 1);
	Q.push(0);
	memset(dis, 0x80, sizeof dis);
	dis[0] = 0;
	while(!Q.empty()){
		int u = Q.front(); Q.pop();
		for(int i = h[u]; i; i = edge[i].next){
			int v = edge[i].to;
			if(dis[v] < dis[u] + edge[i].dis){
				dis[v] = dis[u] + edge[i].dis;
				if(vis[v] == false)
				    vis[v] = true, Q.push(v);
			}
		}vis[u] = false;
	}
	for(int i = 1; i <= n; i ++)
		ans += dis[i];

	printf("%d\n", ans);
}

int c[maxn];

Edge G[maxn << 2];
int H[maxn], CNT;
void Ins(int u, int v){
	CNT ++;
	G[CNT].to = v;
	G[CNT].next = H[u];
	H[u] = CNT;
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++)
	    scanf("%d", &c[i]);
	int u, v, m;
	scanf("%d", &m);
	for(int i = 1; i <= m; i ++){
		scanf("%d%d", &u, &v);
		Ins(u, v); Ins(v, u);
	}

	for(int i = 1; i <= n; i ++){
		for(int j = H[i]; j; j = G[j].next){
			int v = G[j].to;
			if(c[i] < c[v])
			    add(i, v, 1);
			if(c[i] == c[v])
				add(i, v, 0), add(v, i, 0);
			for(int k = H[i]; k; k = G[k].next){
				int u = G[k].to;
				if(c[v] < c[u])
				    add(v, u, 1);
				if(c[v] == c[u])
				    add(v, u, 0), add(u, v, 0);
			}
		}
	}
	Spfa();
	return 0;
}

然而正解是这样的。。对于相邻的人，如果能力值相同那么工资一定相同。这样工资相同的人可能会组成一些强连通分量，我们如果把这些scc缩掉就会形成一个DAG。(这个可以用并查集实现)。跑差分约束(最长路)或者拓扑排序都可。然后对于每一个点相邻的sort一遍，由于相邻的人的的能力值都不同，所以可以在相邻点之间连边即可(建边优化)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#define maxn 100010
using namespace std;

vector<int> G[maxn];

int c[maxn];

bool cmp(int i, int j){return c[i] < c[j];}

struct Edge{
	int to, next;
	Edge(int v = 0, int nxt = 0){
		to = v, next = nxt;
	}
}edge[maxn << 4];
int h[maxn], cnt, n, In[maxn];
void add(int u, int v){
	edge[++ cnt] = Edge(v, h[u]);
	h[u] = cnt;  ++ In[v];
}

int fa[maxn];
int getfa(int x){return x == fa[x] ? x : fa[x] = getfa(fa[x]);}
inline void Union(int x, int y){
	x = getfa(x), y = getfa(y);
	fa[x] = y;
}

stack<int> S;

int ans[maxn], ct[maxn];

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++)
	    scanf("%d", &c[i]), fa[i] = i;
	int u, v, m;
	scanf("%d", &m);
	for(int i = 1; i <= m; i ++){
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
		if(c[u] == c[v])
		    Union(u, v);
	}

	for(int i = 1; i <= n; i ++){
		sort(G[i].begin(), G[i].end(), cmp);
		for(int j = 1; j < G[i].size(); j ++)
			if(c[G[i][j]] == c[G[i][j-1]])
			    Union(G[i][j], G[i][j-1]);
	}
	
	
	for(int i = 1; i <= n; i ++){
		for(int j = 0; j < G[i].size(); j ++)
		    if(c[i] < c[G[i][j]])
		        add(getfa(i), getfa(G[i][j]));
		for(int j = 1; j < G[i].size(); j ++)
		    if(getfa(G[i][j-1]) != getfa(G[i][j]))
		        add(getfa(G[i][j-1]), getfa(G[i][j]));
	}


	for(int i = 1; i <= n; i ++)
	    if(In[i] == 0 && getfa(i) == i)
			S.push(i);

	for(int i = 1; i <= n; i ++)
	    ct[getfa(i)] ++, ans[i] = 1;
	long long ret = 0;
	while(!S.empty()){
		int u = S.top(); S.pop();
		ret += 1ll * ct[u] * ans[u];
		for(int i = h[u]; i; i = edge[i].next){
			int v = edge[i].to;
			In[v] --;
			ans[v] = max(ans[v], ans[u] + 1);
			if(In[v] == 0) S.push(v);
		}
	}

	printf("%lld\n", ret);
	return 0;
}