标签：type iostream class 距离 通过 VID war pair const
\[
Time Limit: 1 s\quad Memory Limit: 256 MB
\]
直接计算即可。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
int a,b,c,d,e,f;
int main(){
scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f);
int ans1 = min(a, d), ans2 = min({b, c, d});
int ans = e>f ? ans1*e + min({ans2, d-ans1})*f : ans2*f + min({ans1, d-ans2})*e;
printf("%d\n", ans);
}\[
Time Limit: 2 s\quad Memory Limit: 256 MB
\]
最后肯定全是 \(W\) 或者全是 \(B\) 的，那么直接暴力扫两遍就可以了。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
char s[maxn];
char ss[maxn];
vector<int> ans;
bool solve() {
ans.clear();
for(int i=1; i<=n; i++) ss[i] = s[i];
for(int i=1; i<n; i++) {
if(ss[i] == 'B') continue;
ss[i] = 'B';
ss[i+1] = s[i+1]=='B' ? 'W':'B';
ans.pb(i);
}
if(ans.size() <= 3*n && ss[n] == 'B') return true;
return false;
}
int main() {
// freopen("in", "r", stdin);
scanf("%d", &n);
scanf("%s", s+1);
if(!solve()) {
for(int i=1; i<=n; i++) {
s[i] = s[i]=='B' ? 'W':'B';
}
if(!solve()) {
return 0*puts("-1");
}
}
int sz = ans.size();
printf("%d\n", sz);
for(int i=0; i<sz; i++) printf("%d%c", ans[i], i==sz-1 ? '\n':' ');
return 0;
}\[
Time Limit: 1 s\quad Memory Limit: 256 MB
\]
假设学校处于 \((sx, sy)\) 位置，可以看成 \((sx, sy)\) 为原点，其他学生到这里的最近距离，那么容易发现，最好的一定可以选成 \((sx-1, sy)、(sx+1, sy)、(sx, sy-1)、(sx, sy+1)\) 这四个之一，所以只要暴力处理这四个点在不在每个学生的最短路上即可。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 2e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
ll sx, sy;
ll x, y;
struct Node {
ll x, y, st;
bool operator < (Node a) const{
return st>a.st;
}
} p[5];
bool ok(Node t) {
ll ans = abs(sx-x) + abs(sy-y);
ll ans1 = abs(sx-t.x) + abs(sy-t.y);
ll ans2 = abs(x-t.x) + abs(y-t.y);
return ans == ans1+ans2;
}
int main() {
// freopen("in", "r", stdin);
scanf("%lld%lld%lld", &n, &sx, &sy);
p[1] = {sx-1, sy, 0};
p[2] = {sx+1, sy, 0};
p[3] = {sx, sy-1, 0};
p[4] = {sx, sy+1, 0};
for(int i=1; i<=n; i++) {
scanf("%lld%lld", &x, &y);
for(int j=1; j<=4; j++) {
if(ok(p[j])) p[j].st++;
}
}
sort(p+1, p+1+4);
printf("%lld\n%lld %lld\n", p[1].st, p[1].x, p[1].y);
return 0;
}\[
Time Limit: 2 s\quad Memory Limit: 256 MB
\]
对于一个 \(v\) 点，如果有多条 \(u->v\)，那么我完全可以到最大的 \(u\) 的时候在派士兵过来，这样做一定不会使结果更劣，并且能够保证尽量占领多的城市。
那么用 \(dp[i][j]\) 表示目前的到达第 \(i\) 个城市，手上有 \(j\) 个士兵时，可以获得的最大价值。
贪心处理 \(j\)，如果有多个城市的最大的 \(u\) 为 \(i\)，既然每个城市都只需派一个人，那么我肯定先派人去可以获得价值大的城市。
view
#include <map>
#include <set>
#include <list>
#include <tuple>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
#define INOPEN freopen("in.txt", "r", stdin)
#define OUTOPEN freopen("out.txt", "w", stdout)
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 5e3 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m, k;
int cas, tol, T;
vector<int> g[maxn];
int a[maxn], b[maxn], c[maxn], p[maxn];
ll dp[2][maxn];
bool cmp(int x, int y) {
return c[x] > c[y];
}
int main() {
scanf("%d%d%d", &n, &m, &k);
for(int i=1, x, y, z; i<=n; i++) {
scanf("%d%d%d", &a[i], &b[i], &c[i]);
p[i] = i, g[i].clear();
}
for(int i=1, u, v; i<=m; i++) {
scanf("%d%d", &u, &v);
p[v] = max(p[v], u);
}
for(int i=1; i<=n; i++) g[p[i]].pb(i);
for(int i=1; i<=n; i++) sort(g[i].begin(), g[i].end(), cmp);
for(int i=0; i<=5000; i++)dp[0][i] = dp[1][i] = -INF;
dp[0][k] = 0;
int f = 0;
for(int i=1; i<=n; i++) {
f = !f;
for(int j=0; j<=5000; j++) dp[f][j] = -INF;
for(int j=a[i]; j<=5000; j++) {
if(dp[!f][j] == -INF) continue;
ll w = j+b[i], d = 0;
dp[f][w] = max(dp[f][w], dp[!f][j]);
for(auto k : g[i]) {
d += c[k], w--;
dp[f][w] = max(dp[f][w], dp[!f][j]+d);
if(!w) break;
}
}
}
ll ans = -1;
for(int i=0; i<=5000; i++) ans = max(ans, dp[f][i]);
printf("%lld\n", ans);
return 0;
}\[
Time Limit: 2 s\quad Memory Limit: 256 MB
\]
对于一个数 \(x\)
当 \(x\) 的奇偶性相同时，显然对于同一个上届，\(x\) 越小，满足条件的数字越多，那么就可以对 \(x\) 奇偶然后二分答案，最后取两者较大值。
view
#include <map>
#include <set>
#include <list>
#include <tuple>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<ll, ll>
#define INOPEN freopen("in.txt", "r", stdin)
#define OUTOPEN freopen("out.txt", "w", stdout)
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
ll n, m;
int cas, tol, T;
bool ok(ll x) {
queue<pii> q;
if(x&1) q.push({x, x});
else q.push({x, x+1});
ll ans = 0;
while(!q.empty()) {
auto it = q.front();
q.pop();
ans += min(n, it.se)-it.fi+1;
if(2ll*it.fi <= n) q.push({2ll*it.fi, 2ll*it.se+1});
}
return ans>=m;
}
int main() {
scanf("%lld%lld", &n, &m);
ll ans = 0;
{
ll l=0, r=n/2, res=0;
while(l<=r) {
ll mid = l+r>>1;
if(ok(mid<<1|1)) l = mid+1, res=mid;
else r = mid-1;
}
ans = max(ans, res<<1|1);
}
{
ll l=1, r=n/2, res=0;
while(l<=r) {
ll mid = l+r>>1;
if(ok(mid<<1)) l = mid+1, res=mid;
else r = mid-1;
}
ans = max(ans, res<<1);
}
printf("%lld\n", ans);
return 0;
}\[
Time Limit: 8 s\quad Memory Limit: 256 MB
\]
留坑
Codeforces Round #608 (Div. 2) 题解
标签：type iostream class 距离 通过 VID war pair const
原文地址：https://www.cnblogs.com/Jiaaaaaaaqi/p/12063536.html
