称检测点查询
实现
#include<bits/stdc++.h>
using namespace std;
int n,X,Y;
const int N = 210;
struct node
{
int x,y;
int d;
int id;
bool operator< (const node& t) const
{
if(d!=t.d)
{
return d<t.d;
}
else return id<t.id;
}
}nodes[N];
int main()
{
cin>>n>>X>>Y;
int a,b;
for(int i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
nodes[i].x = a;
nodes[i].y = b;
nodes[i].id = i+1;
nodes[i].d = (a-X)*(a-X)+(b-Y)*(b-Y);
}
sort(nodes,nodes+n);
for(int i=0;i<3;i++) cout<<nodes[i].id<<endl;
}
风险人群筛查
实现
#include<iostream>
#include<map>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
int n, k, t, x1, y1, x2, y2;
int main()
{
cin >> n >> k >> t >> x1 >> y1 >> x2 >> y2;
int res1 = 0, res2 = 0;
while (n -- )
{
bool r1 = 0, r2 = 0;
int s = 0;
for (int i = 0; i < t; i ++ )
{
int x, y;
cin >> x >> y;
if (x >= x1 && x <= x2 && y >= y1 && y <= y2)
{
s ++ ;
r1 = true;
if (s >= k) r2 = true;
}
else s = 0;
}
if (r1) res1 ++ ;
if (r2) res2 ++ ;
}
cout << res1 << endl << res2 << endl;
return 0;
}
点亮数字人生
思路
来自acwing-yxc https://www.acwing.com/activity/content/code/content/935832/
- 判断环和计算的顺序需要拓扑排序
- 需要将输入值和输出器件的编号读入vector中
实现
#include<bits/stdc++.h>
using namespace std;
const int N = 3010, M = N * 5;
int m, n;
int w[N], f[N];
int h[N], e[M], ne[M], idx;
int q[N], d[N];
vector<int> in[M], out[M];
int get(char* str)
{
string names[] = {
"AND", "OR", "NOT", "XOR", "NAND", "NOR"
};
for (int i = 0; i < 6; i ++ )
if (names[i] == str)
return i;
return -1;
}
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
d[b] ++ ;
}
bool topsort() //数组q中保存拓扑排序序列
{
int hh = 0, tt = -1;
for (int i = 1; i <= m + n; i ++ )
if (!d[i])
q[ ++ tt] = i;
while (hh <= tt)
{
int t = q[hh ++ ];
for (int i = h[t]; ~i; i = ne[i])
{
int j = e[i];
if ( -- d[j] == 0)
q[ ++ tt] = j;
}
}
return tt == m + n - 1;
}
int main()
{
int T;
scanf("%d", &T);
while (T -- )
{
scanf("%d%d", &m, &n);
memset(h, -1, sizeof h);
idx = 0;
memset(d, 0, sizeof d);
char str[100];
for (int i = 1; i <= n; i ++ )
{
int cnt;
scanf("%s%d", str, &cnt);
f[m + i] = get(str);
while (cnt -- )
{
scanf("%s", str);
int t = atoi(str + 1);
if (str[0] == 'I') add(t, m + i);
else add(m + t, m + i);
}
}
int Q;
scanf("%d", &Q);
for (int i = 0; i < Q; i ++ )
{
in[i].clear();
for (int j = 0; j < m; j ++ )
{
int x;
scanf("%d", &x);
in[i].push_back(x);
}
}
for (int i = 0; i < Q; i ++ )
{
out[i].clear();
int cnt;
scanf("%d", &cnt);
while (cnt -- )
{
int x;
scanf("%d", &x);
out[i].push_back(x);
}
}
if (!topsort()) puts("LOOP");
else
{
for (int i = 0; i < Q; i ++ )
{
for (int j = 0; j < m; j ++ ) w[j + 1] = in[i][j];
for (int j = m + 1; j <= m + n; j ++ )
if (f[j] == 0 || f[j] == 5) w[j] = 1; //AND和NOR时需要把初值设置为1
else w[j] = 0;
for (int j = 0; j < m + n; j ++ )
{
int t = q[j], v = w[t]; //从q中取数，按照拓扑排序计算
for (int k = h[t]; ~k; k = ne[k])
{
int u = e[k];
if (f[u] == 0) w[u] &= v;
else if (f[u] == 1) w[u] |= v;
else if (f[u] == 2) w[u] = !v;
else if (f[u] == 3) w[u] ^= v;
else if (f[u] == 4) w[u] |= !v;
else w[u] &= !v;
}
}
for (auto x: out[i])
printf("%d ", w[m + x]);
puts("");
}
}
}
return 0;
}