CSP202203 全部题解
作者：互联网
都是些简单题，但我看网上好像没有全题解，写一篇造福大学生。
未初始化警告
签到
#include <iostream>
#include <cstdio>
using namespace std;
#define N 100005
int n, k, ans = 0;
bool vis[N];
int main() {
ios::sync_with_stdio(false);
cin >> n >> k;
vis[0] = true;
for(int i = 1; i <= k; ++i) {
int x, y;
cin >> x >> y;
if(!vis[y]) ++ans;
vis[x] = true;
}
cout << ans;
return 0;
}
出行计划
签到
#include <iostream>
#include <cstdio>
using namespace std;
#define N 400005
int n, m, k;
int d[N];
int main() {
ios::sync_with_stdio(false);
cin >> n >> m >> k;
for(int i = 1; i <= n; ++i) {
int t, c;
cin >> t >> c;
int left = max(1, t - c + 1);
++d[left], --d[t + 1];
}
for(int i = 1; i <= N - 5; ++i) d[i] += d[i - 1];
for(int i = 1; i <= m; ++i) {
int q;
cin >> q;
cout << d[q + k] << endl;
}
return 0;
}
计算资源调度器
简单模拟
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
#define N 3005
int n, m, g;
int f[N], a[N], na[N], pa[N], paa[N];
bool paar[N];
struct node {
int cnt, l, id;
bool operator <(const node &x) const {
return cnt == x.cnt ? id < x.id : cnt < x.cnt;
}
}nd[N], p[N];
vector < int > seg[N], task[N];
bool used[N], tmp[N];
void nodeFrd(int na) {
for(int i = 0; i < seg[na].size(); ++i) {
int u = seg[na][i];
used[u] = false;
}
for(int i = 1; i <= n; ++i) used[i] ^= 1;
}
void taskFrd(int pa) {
for(int i = 1; i <= m; ++i) {
bool flag = false;
for(int j = 0; j < seg[i].size(); ++j) {
int u = seg[i][j];
for(int k = 0; k < task[u].size(); ++k) {
if(a[task[u][k]] == pa) flag = true;
}
}
if(!flag) {
for(int j = 0; j < seg[i].size(); ++j) used[seg[i][j]] = false;
}
}
}
void taskUFrd(int paa) {
for(int i = 1; i <= n; ++i) {
bool flag = true;
for(int j = 0; j < task[i].size(); ++j) {
if(a[task[i][j]] == paa) flag = false;
}
if(!flag) used[i] = false;
}
}
bool check() {
bool res = false;
for(int i = 1; i <= n; ++i) res |= used[i];
return res;
}
int getans() {
int tot = 0;
for(int i = 1; i <= n; ++i) if(used[i]) p[++tot] = nd[i];
sort(p + 1, p + tot + 1);
return p[1].id;
}
/*调试内容
void print() {
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < task[i].size(); ++j) {
cout << task[i][j] << ' ';
}
cout << endl;
}
cout << "----------\n";
}
*/
int main() {
ios::sync_with_stdio(false);
cin >> n >> m;
for(int i = 1; i <= n; ++i) {
cin >> nd[i].l;
nd[i].id = i;
seg[nd[i].l].push_back(i);
}
cin >> g;
for(int Tc = 1; Tc <= g; ++Tc) {
cin >> f[Tc] >> a[Tc] >> na[Tc] >> pa[Tc] >> paa[Tc] >> paar[Tc];
for(int i = 1; i <= f[Tc]; ++i) {
fill(used + 1, used + n + 1, true);
if(na[Tc]) {
nodeFrd(na[Tc]);
}
if(pa[Tc]) {
taskFrd(pa[Tc]);
}
if(paa[Tc]) {
if(paar[Tc]) {
taskUFrd(paa[Tc]);
} else {
for(int j = 1; j <= n; ++j) tmp[j] = used[j];
taskUFrd(paa[Tc]);
if(!check()) {
for(int j = 1; j <= n; ++j) used[j] = tmp[j];
}
}
}
if(!check()) cout << 0 << ' ';
else {
int ans = getans();
++nd[ans].cnt;
task[ans].push_back(Tc);
cout << ans << ' ';
}
}
cout << endl;
}
return 0;
}
通信系统管理
简单数据结构。用 set 维护边即可。
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <set>
using namespace std;
#define int long long
#define N 200005
#define Prr pair < int, int >
#define fi first
#define se second
int n, m;
int k[N >> 1], l[N >> 1];
int tot1 = 0, u[N], v[N], x[N], y[N], t[N];
int tot2 = 0, num[N];
bool p[N], q[N];
vector < int > e[N], val[N], M[N];
void init() {
cin >> n >> m;
for(int i = 1; i <= m; ++i) {
cin >> k[i];
for(int j = 1; j <= k[i]; ++j) {
++tot1;
cin >> u[tot1] >> v[tot1] >> x[tot1] >> y[tot1];
t[tot1] = i + y[tot1];
e[u[tot1]].push_back(v[tot1]);
e[v[tot1]].push_back(u[tot1]);
val[u[tot1]].push_back(0);
val[v[tot1]].push_back(0);
}
cin >> l[i];
for(int j = 1; j <= l[i]; ++j) cin >> num[++tot2];
cin >> p[i] >> q[i];
}
}
struct cmp1 {
bool operator ()(Prr a, Prr b) {
return a.se == b.se ? a.fi > b.fi : a.se < b.se;
}
};
set < Prr, cmp1 > Val[N];
int ans[N], cntP = 0, cntQ = 0;
void updPr(int u, int v, int x) {
if(ans[u] == v && ans[v] == u) cntQ += x;
else {
if(ans[ans[u]] == u) cntQ += x;
if(ans[ans[v]] == v) cntQ += x;
}
}
void mod(int u, int v, int x) {
int p = lower_bound(e[u].begin(), e[u].end(), v) - e[u].begin();
Val[u].erase(make_pair(e[u][p], val[u][p]));
val[u][p] += x;
if(val[u][p] > 0) Val[u].insert(make_pair(e[u][p], val[u][p]));
if(!ans[u]) --cntP;
ans[u] = !Val[u].empty() ? (*Val[u].rbegin()).fi : 0;
if(!ans[u]) ++cntP;
}
void upd(int tm) {
for(int i = 0; i < M[tm].size(); ++i) {
int p = M[tm][i];
updPr(u[p], v[p], -1);
mod(u[p], v[p], -x[p]);
mod(v[p], u[p], -x[p]);
updPr(u[p], v[p], 1);
}
}
void solve() {
for(int i = 1; i <= n; ++i) sort(e[i].begin(), e[i].end());
cntP = n;
int pos1 = 0, pos2 = 0;
for(int i = 1; i <= m; ++i) {
upd(i);
for(int j = pos1 + 1; j <= pos1 + k[i]; ++j) {
updPr(u[j], v[j], -1);
mod(u[j], v[j], x[j]);
mod(v[j], u[j], x[j]);
updPr(u[j], v[j], 1);
M[t[j]].push_back(j);
}
pos1 += k[i];
for(int j = pos2 + 1; j <= pos2 + l[i]; ++j) {
cout << ans[num[j]] << endl;
}
pos2 += l[i];
if(p[i]) cout << cntP << endl;
if(q[i]) cout << cntQ << endl;
}
}
signed main() {
init();
solve();
return 0;
}
博弈论与石子合并
简单博弈
#include <iostream>
#include <cstdio>
using namespace std;
#define N 100005
int n, a[N];
bool k;
void subtask1() {
int res = 0, SIZE = (n >> 1) + 1, ans = 1e9;
for(int i = 1; i <= n; ++i) {
res += a[i];
if(i >= SIZE) res -= a[i - SIZE], ans = min(ans, res);
}
cout << ans << endl;
}
bool check(int x) {
int res = 0, cnt = 0;
for(int i = 1; i <= n; ++i) {
res += a[i];
if(res >= x) res = 0, ++cnt;
}
return cnt > n / 2;
}
void subtask2() {
int l = 1, r = 1e9, mid, ans = 0;
while(l <= r) {
mid = l + r >> 1;
if(check(mid)) l = mid + 1, ans = mid;
else r = mid - 1;
}
cout << ans << endl;
}
int main() {
ios::sync_with_stdio(false);
cin >> n >> k;
for(int i = 1; i <= n; ++i) cin >> a[i];
if((n & 1) ^ k) subtask1();
else subtask2();
return 0;
}
标签：CSP202203,int,题解,tot1,全部,ans,include,Tc,define 来源： https://www.cnblogs.com/zhouyuhang114/p/16377074.html