A - New Building for SIS
题意:
一排房子1−n1−n,问从第xx个结点到第个结点 需要的最少步数 (只能在特定的楼层内转移)
题解
判断几个就可以了
trick点 : 注意特定的楼层内部也是需要计算步数
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
void F() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
}
int main() {
F();
int n, h, a, b, k;
cin >> n >> h >> a >> b >> k;
while(k--) {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
int ans = 0;
if(x2 - x1 == 0)
ans += abs(y2-y1);
else if(y1 > b)
ans += abs(y1-b+abs(x2-x1)+abs(y2-b));
else if(y1 < a)
ans += abs(a-y1+abs(x1-x2)+abs(y2-a));
else
ans += abs(abs(x2-x1)+abs(y2-y1));
cout << ans << endl;
}
return 0;
}
B - Badge
题意:
一个图,给出每个点到另一个点的关系, 问从一个点 第二次到另一个点是哪个点
题解:
直接对每个点dfs即可, 没有任何trick点
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define ls o<<1
#define rs o<<1|1
#define fi first
#define se second
#define pii pair<int ,int>
#define pll pair<ll, ll>
#define CLR(a, b) memset(a, (b), sizeof(a))
const int INF = 0x3f3f3f3f;
const ll INFLL = ~0ull>>1;
const int mod = 1e9+7;
const int MAXN = (int)3e3+10;
void F() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
}
vector<int> v[MAXN], vv;
bool vis[MAXN];
int ans;
void dfs(int u) {
for(auto v: v[u]) {
if(vis[v]) {
vv.pb(v); return ;
}
vis[v] = true;
dfs(v);
}
}
int main(int argc, char const *argv[])
{
F();
int n, x;
cin >> n;
for(int i = 1; i <= n ; ++i) {
cin >> x;
v[i].pb(x);
}
for(int i = 1; i <= n; ++i) {
CLR(vis, false);
vis[i] = true;
dfs(i);
}
for(auto i : vv) {
cout << i << " ";
}
return 0;
}
C - Elections
题意:
nn个投票人, 个被投票人, 给出这nn个人会把票投给谁,并且收买他的代价, 最终求得这个投票人票数最大的代价最小是多少
题解:
O(n∗m)O(n∗m) 的复杂度暴力即可, 维护一个优先队列, 每次枚举要收买ii个人的总代价,
trick: 不要忘记了初始对每个人排序
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define ls o<<1
#define rs o<<1|1
#define fi first
#define se second
#define pii pair<int ,int>
#define pll pair<ll, ll>
#define CLR(a, b) memset(a, (b), sizeof(a))
const int INF = 0x3f3f3f3f;
const ll INFLL = ~0ull>>1;
const int mod = 1e9+7;
const int MAXN = (int)3e3+10;
void F() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
}
vector<int> v[MAXN];
int main() {
F();
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; ++i) {
int id, x;
cin >> id >> x;
v[id].pb(x);
}
for(int i = 2; i <= m; ++i)
sort(v[i].begin(), v[i].end());
ll ans = INFLL;
priority_queue<int, vector<int>, greater<int> > que;
for(int i = 0; i < n; ++i) {
ll tmp = 0;
int l = (int)v[1].size();
while(!que.empty()) que.pop();
for(int j = 2; j <= m; ++j) {
int z = (int)v[j].size()-i;
for(int k = 0; k < z; ++k) {
l++; tmp += v[j][k];
}
int zz = max(z, 0);
for(int k = zz; k < (int)v[j].size(); ++k)
que.push(v[j][k]);
}
while(!que.empty() && l<=i) {
l++; tmp += que.top(); que.pop();
}
if(l > i) ans = min(ans, tmp);
}
cout << ans << endl;
return 0;
}