CCPC-Wannafly Summer Camp 2019 Day1
A - Jzzhu and Cities
题目大意：给个图，m条普通边，k条特殊边，保证1到N的最短路上最多可以删掉几条特殊边使得最短路不变。
解法1：跑最短路后松弛一遍所有边，如果松弛时既有普通和特殊则可去掉特殊铁路。
解法2：直接重载搜索顺序使得迪杰斯特拉优先跑普通铁路，再记录总特殊条数-跑的条数即为答案。
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 1e6 + 10;
int n, m, k;
bool used[MAXN];
struct node
{
ll to, val, ok;
node() { }
node(ll a, ll b, ll c) { to = a, val = b, ok = c; }
};
vector <node> edge[MAXN];
struct bfsnode
{
ll now, dis, cnt;
bfsnode() { }
bfsnode(ll a, ll b, ll c) { now = a, dis = b, cnt = c; }
};
bool operator < (bfsnode a, bfsnode b)
{
if(a.dis == b.dis)
return a.cnt > b.cnt;
return a.dis > b.dis;
}
void bfs()
{
priority_queue <bfsnode> PQ;
PQ.push(bfsnode(1, 0, 0));
ll ans = 0;
while(!PQ.empty())
{
bfsnode rnow = PQ.top();
PQ.pop();
if(used[rnow.now])
continue;
used[rnow.now] = true;
if(rnow.cnt)
++ans;
for(auto x : edge[rnow.now])
{
PQ.push(bfsnode(x.to, rnow.dis + x.val, x.ok));
}
}
cout << k - ans << '\n';
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
//freopen("azerah.in","r",stdin);
//freopen("azerah.out","w",stdout);
cin >> n >> m >> k;
for(int i = 0; i < m; ++i)
{
ll a, b, c;
cin >> a >> b >> c;
edge[a].push_back(node(b, c, 0));
edge[b].push_back(node(a, c, 0));
}
for(int i = 0; i < k; ++i)
{
ll a, b;
cin >> a >> b;
edge[1].push_back(node(a, b, 1));
edge[a].push_back(node(1, b, 1));
}
bfs();
return 0;
}
B - Phillip and Trains
人只能向右一步同时向上 或 向下 或不动， 然后火车向左两步，求会不会被撞。
解法：相对运动转化+二维dp，当前位置只能由三个位置转移而来，如果能转移到最右端则不会被撞
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 4e6 + 10;
char arr[5][200];
int ok[5][200];
bool check(int x, int y)
{
if ( arr[x][y] != '.' || arr[x][y - 1] != '.' || arr[x][y - 2] != '.')
return false;
if(arr[x][y - 3] == '.' && ok[x][y - 3])
return true;
if(x != 2)
{
if(arr[x + 1][y - 2] == '.' && arr[x + 1][y - 3] == '.' && ok[x + 1][y - 3])
return true;
}
if(x != 0)
{
if(arr[x - 1][y - 2] == '.' && arr[x - 1][y - 3] == '.' && ok[x - 1][y - 3])
return true;
}
return false;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int t;
cin >> t;
while(t--)
{
memset(arr, '.', sizeof(arr));
memset(ok, 0, sizeof(ok));
int n, k;
cin >> n >> k;
for(int i = 0; i < 3; ++i)
{
for(int j = 0; j < n; ++j)
{
cin >> arr[i][j];
if(arr[i][j] == 's')
ok[i][j] = 1, arr[i][j] = '.';
}
}
bool flag = false;
for(int j = 3; j < n + 10; j++)
{
for(int i = 0; i < 3; ++i)
{
if(check(i, j))
{
ok[i][j] = true;
if(j >= n - 1)
flag = true;
}
}
}
// for(int i = 0; i < 3; ++i)
// {
// for(int j = 0; j < n; ++j)
// {
// cout << ok[i][j] ;
// }
// cout << '\n';
// }
if(flag)
cout << "YES\n";
else
cout << "NO\n";
}
return 0;
}
/*
1
16 4
...AAAAA........
s.BBB....CCCCC..
.......DDDDD....
*/C - A Mist of Florescence
题意，构造一个50*50 以内的矩阵，包含a个A联通快，b个B联通块，c个C联通块，d个D联通块。
解法：构造4个25*25的ABCD矩阵连在一起， 然后每九宫格往里填即可。
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 1e2 + 10;
char mp[MAXN][MAXN] = {0};
void init(int x, int y, char c)
{
for(int i = x; i < x + 25; ++i)
{
for(int j = y; j < y + 25; ++j)
{
mp[i][j] = c;
}
}
}
void show()
{
for(int i = 0; i < 50; ++i)
{
for(int j = 0; j < 50; ++j)
{
cout << mp[i][j];
}
cout << '\n';
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
init(0, 0, 'A');
init(0, 25, 'B');
init(25, 0, 'C');
init(25, 25, 'D');
--a, --b, --c, --d;
for(int i = 1; i < 50; i += 2)
{
if(i == 25)
++i;
for(int j = 1; j < 50; j += 2)
{
if(j == 25)
++j;
if(a && mp[i][j] != 'A')
--a, mp[i][j] = 'A';
else if(b && mp[i][j] != 'B')
--b, mp[i][j] = 'B';
else if(c && mp[i][j] != 'C')
--c, mp[i][j] = 'C';
else if(d && mp[i][j] != 'D')
--d, mp[i][j] = 'D';
}
}
cout << "50 50\n";
show();
return 0;
} D - Unbearable Controversy of Being
题意：求图中菱形个数
遍历所有点能达到的长度为2的路径的到达点，map计数，加法原理记录答案。
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double ld;
typedef std::pair<int, int> pii;
#define mp(x, y) make_pair(x, y)
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 1e18;
const ll MOD = 1e9 + 7;
const int MAXN = 3e3 + 10;
int n, m;
vector <int> edge[MAXN];
int ans = 0;
map <int, int> rec;
bitset <MAXN> used[MAXN];
void cal(int now)
{
rec.clear();
for(int i = 1; i <= n; ++i)
used[i].reset();
for(auto x : edge[now])
{
for(auto y : edge[x])
{
if(!used[x][y] && y != now)
{
ans += rec[y];
++rec[y];
}
}
}
return ;
}
int main()
{
cin >> n >> m;
for(int i = 0, from, to; i < m; ++i)
cin >> from >> to, edge[from].push_back(to);
for(int i = 1; i <= n; ++i)
{
cal(i);
}
cout << ans << '\n';
return 0;
}E - Igor In the Museum
题意：每个.和*之间都有一副画，求点x，y所在联通块内画的总数。
思路：dfs染色记录答案输出
我用的二维映射一维＋并查集
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 2e6 + 10;
char mp[1100][1100];
bool used[1100][1100][4];
int fa[MAXN] = {0};
int val[MAXN] = {0};
int movl[4][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int findfa(int x)
{
return x == fa[x] ? x : fa[x] = findfa(fa[x]);
}
int n, m, k;
int tra(int x, int y)
{
return (x - 1) * m + y;
}
void oprt(int x, int y)
{
for(int i = 0; i < 4; ++i)
{
int rx = x + movl[i][0], ry = y + movl[i][1], rp = tra(x, y);
if(mp[rx][ry] == '*' && !used[rx][ry][i])
{
++val[rp];
used[rx][ry][i] = true;
}
}
for(int i = 0; i < 2; ++i)
{
int rx = x + movl[i][0], ry = y + movl[i][1], rp = tra(rx, ry), rrp = tra(x, y);
if(mp[rx][ry] == '.')
{
int p = findfa(rp), q = findfa(rrp);
if(p != q)
{
fa[p] = q;
val[q] += val[p];
}
}
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n >> m >> k;
for(int i = 0; i <= n * m + n + m + 100; ++i)
{
fa[i] = i;
}
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)
{
cin >> mp[i][j];
}
}
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)
{
if(mp[i][j] == '.')
{
oprt(i, j);
}
}
}
while(k--)
{
int x, y;
cin >> x >> y;
int rp = findfa( tra(x, y) );
cout << val[rp] << '\n';
}
return 0;
} F - The Child and Toy
智障题，删每条边两端点最小点权之和
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 1e6 + 10;
ll val[MAXN];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; ++i)
{
cin >> val[i];
}
ll ans = 0;
for(int i = 1; i <= m; ++i)
{
int p, q;
cin >> p >> q;
ans += min(val[p], val[q]);
}
cout << ans << '\n';
return 0;
} G - New Year Permutation
一个矩阵，若i,j == j,i == 1则这部分有传递性可互换，求最小排列。
用并查集维护联通块后set排序+位置直接复写数组输出即可
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double ld;
typedef std::pair<int, int> pii;
#define mp(x, y) make_pair(x, y)
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 1e18;
const ll MOD = 1e9 + 7;
const int MAXN = 5e2 + 10;
int fa[MAXN];
int findfa(int x)
{
return x == fa[x] ? x : fa[x] = findfa(fa[x]);
}
int arr[MAXN] = {0};
bitset <MAXN> ok[MAXN];
set <int> ST[MAXN], RT[MAXN];
int main()
{
int n;
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> arr[i];
for(int i = 1; i <= n; ++i)
{
fa[i] = i;
for(int j = 1; j <= n; ++j)
{
char x;
cin >> x;
ok[i][j] = x == '1';
}
}
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= n; ++j)
{
if(ok[i][j] && ok[j][i])
{
int p = findfa(i), q = findfa(j);
if(p != q)
fa[p] = q;
}
}
}
for(int i = 1; i <= n; ++i)
{
int p = findfa(i);
ST[p].insert(arr[i]);
RT[p].insert(i);
}
for(int i = 1; i <= n; ++i)
{
auto r = RT[i].begin();
for(auto x : ST[i])
{
arr[ *r ] = x;
++r;
}
}
for(int i = 1; i <= n; ++i)
cout << arr[i] << ' ';
return 0;
}H - Alyona and the Tree
规定1为根节点，若存在dis（u，v） > val(u)则称u 伤心，问需要删多少点才让整棵树不伤心
从根dfs统计不伤心的点，加上一点的树上dp （最大路径和）即可
/*
Zeolim - An AC a day keeps the bug away
*/
//pragma GCC optimize(2)
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef std::pair<int, int> pii;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
const int MAXN = 1e6 + 10;
ll val[MAXN] = {0};
bool used[MAXN] = {0};
struct node
{
ll to, val;
node(){}
node(int x, int y)
{
to = x;
val = y;
}
};
vector <node> edge[MAXN];
int n, k;
void dfs(ll now, ll sum)
{
if(sum > val[now])
return;
used[now] = true;
for(int i = 0; i < edge[now].size(); ++i)
{
if(!used[edge[now][i].to])
dfs(edge[now][i].to, max(sum + edge[now][i].val, edge[now][i].val));
}
}
int main()
{
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> val[i];
for(int i = 2; i <= n; ++i)
{
int to, val;
cin >> to >> val;
edge[i].push_back(node(to, val));
edge[to].push_back(node(i, val));
}
dfs(1, 0);
int ans = n;
for(int i = 1; i <= n; ++i)
{
if(used[i])
--ans;
}
cout << ans << '\n';
return 0;
}
