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;
}