Codeforces Round #684 (Div. 2)
贪心模拟
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int _, c0, c1, h, n; char s[N]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif _ = read(); while (_--) { n = read(); c0 = read(); c1 = read(); h = read(); scanf("%s", s + 1); int ans = 0; for (int i = 1; i <= n; ++i) { if (s[i] == '0') ans += min(c0, c1 + h); else ans += min(c1, c0 + h); } cout << ans << ' '; } // #ifndef ONLINE_JUDGE // system("pause"); // #endif }
发现一定是从最后开始取
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 1e6 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int _, n, k; int a[N]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif _ = read(); while (_--) { n = read(); k = read(); rep(i, 1, n * k) a[i] = read(); ll sum = 0; ll now = n * k - n / 2, cnt = 1; while (cnt <= k) { sum += a[now]; now -= n / 2 + 1; cnt++; } printf("%lld ",sum); } // #ifndef ONLINE_JUDGE // system("pause"); // #endif }
C1 - Binary Table (Easy Version)
简单版本的话对于每个1都可以通过三次操作让其变成0
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 210, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int _, n, m, cnt; char s[N][N]; struct node { int x1, y1, x2, y2, x3, y3; }e[M]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif _ = read(); while (_--) { cnt = 0; n = read(); m = read(); rep(i, 1, n) scanf("%s", s[i] + 1); rep(i, 1, n) { rep(j, 1, m) { if (s[i][j] == '0') continue; if (i == 1 && j == 1) { e[++cnt] = {1, 1, 1, 2, 2, 1}; e[++cnt] = {1, 1, 1, 2, 2, 2}; e[++cnt] = {1, 1, 2, 1, 2, 2}; continue; } if (i == n && j == m) { e[++cnt] = {n, m, n, m - 1, n - 1, m}; e[++cnt] = {n, m, n, m - 1, n - 1, m - 1}; e[++cnt] = {n, m, n - 1, m - 1, n - 1, m}; continue; } if (i + 1 <= n && j + 1 <= m) { e[++cnt] = {i,j,i+1,j,i+1,j+1}; e[++cnt] = {i,j,i,j+1,i+1,j+1}; e[++cnt] = {i,j,i+1,j,i,j+1}; } else if (i > 1 && j > 1) { e[++cnt] = {i,j,i,j-1,i-1,j-1}; e[++cnt] = {i,j,i,j-1,i-1,j}; e[++cnt] = {i,j,i-1,j-1,i-1,j}; } else if (i + 1 <= n && j - 1 >= 1) { e[++cnt] = {i,j,i+1,j-1,i,j-1}; e[++cnt] = {i,j,i+1,j-1,i+1,j}; e[++cnt] = {i,j,i+1,j,i,j-1}; } else { e[++cnt] = {i,j,i-1,j,i,j+1}; e[++cnt] = {i,j,i-1,j,i-1,j+1}; e[++cnt] = {i,j,i,j+1,i-1,j+1}; } } } cout << cnt << ' '; rep(i, 1, cnt) { cout << e[i].x1 << " " << e[i].y1 << " " << e[i].x2 << " " << e[i].y2 << " " << e[i].x3 << " " << e[i].y3 << ' '; } } // #ifndef ONLINE_JUDGE // system("pause"); // #endif }
C2 - Binary Table (Hard Version)
hard版本可以发现对于每一个2*2的矩形我们都可以在四步之内将其全置成0
那么我们只需判断行和列是否为奇数，若为奇数预处理掉多出来的一行即可
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 210, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int _, n, m, cnt; char s[N][N]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } struct node { int x1, y1, x2, y2, x3, y3; }e[M]; void f(int x, int y) { if (s[x][y] == '1') s[x][y] = '0'; else s[x][y] = '1'; } void f1(PII x) { f(x.fi, x.se); f(x.fi - 1, x.se); f(x.fi, x.se + 1); e[++cnt] = {x.fi, x.se, x.fi - 1, x.se, x.fi, x.se + 1}; } void f2(PII x) { f(x.fi, x.se); f(x.fi + 1, x.se); f(x.fi, x.se + 1); e[++cnt] = {x.fi, x.se, x.fi + 1, x.se, x.fi, x.se + 1}; } void f3(PII x) { f(x.fi, x.se); f(x.fi + 1, x.se); f(x.fi, x.se - 1); e[++cnt] = {x.fi, x.se, x.fi + 1, x.se, x.fi, x.se - 1}; } void f4(PII x) { f(x.fi, x.se); f(x.fi - 1, x.se); f(x.fi, x.se - 1); e[++cnt] = {x.fi, x.se, x.fi - 1, x.se, x.fi, x.se - 1}; } void f1(int x, int y) {f1({x, y});} void f2(int x, int y) {f2({x, y});} void f3(int x, int y) {f3({x, y});} void f4(int x, int y) {f4({x, y});} int get(int x, int y) { int sum = 0; sum += s[x][y] == '1'; sum += s[x][y + 1] == '1'; sum += s[x + 1][y] == '1'; sum += s[x + 1][y + 1] == '1'; return sum; } void solve(int x, int y) { if (get(x, y) == 1) { if (s[x][y] == '1') { f2(x, y); f3(x, y + 1); f1(x + 1, y); return ; } if (s[x + 1][y] == '1') { f1(x + 1, y); f2(x, y); f4(x + 1, y + 1); return ; } if (s[x][y + 1] == '1') { f3(x, y + 1); f2(x, y); f4(x + 1, y + 1); return ; } if (s[x + 1][y + 1] == '1') { f4(x + 1, y + 1); f3(x, y + 1); f1(x + 1, y); return ; } } else if (get(x, y) == 2) { int ans = 0; if (s[x][y] == '1' && s[x + 1][y + 1] == '1') { // puts("1!"); f2(x, y); f4(x + 1, y + 1); return ; } if (s[x][y + 1] == '1' && s[x + 1][y] == '1') { // puts("2!"); f1(x + 1, y); f3(x, y + 1); return ; } if (s[x][y] == '1' && s[x][y + 1] == '1') { // 1100 // puts("3!"); f2(x, y); f1(x + 1, y); f2(x, y); f4(x + 1, y + 1); return ; } if (s[x + 1][y] == '1' && s[x + 1][y + 1] == '1') { // 0011 // puts("4!"); f1(x + 1, y); f2(x, y); f3(x, y + 1); f1(x + 1, y); return ; } if (s[x][y] == '1' && s[x + 1][y] == '1') { // puts("5!"); f2(x, y); f3(x, y + 1); f2(x, y); f4(x + 1, y + 1); return ; } if (s[x][y + 1] == '1' && s[x + 1][y + 1] == '1') { // puts("6!"); f3(x, y + 1); f2(x, y); f3(x, y + 1); f1(x + 1, y); return ; } } else if (get(x, y) == 3) { if (s[x][y] == '0') e[++cnt] = {x, y + 1, x + 1, y, x + 1, y + 1}; else if (s[x][y + 1] == '0') e[++cnt] = {x, y, x + 1, y, x + 1, y + 1}; else if (s[x + 1][y] == '0') e[++cnt] = {x + 1, y + 1, x, y, x, y + 1}; else if (s[x + 1][y + 1] == '0') e[++cnt] = {x, y, x + 1, y, x, y + 1}; } else if (get(x, y) == 4) { f4(x + 1, y + 1); f2(x, y); f3(x, y + 1); f1(x + 1, y); } return; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif _ = read(); while (_--) { cnt = 0; n = read(); m = read(); rep(i, 1, n) { scanf("%s", s[i] + 1); } if (n & 1) { rep(i, 1, m) { if (s[1][i] == '0') continue; if (i != m) { f2({1, i}); } else { f4({2, i}); } } } if (m & 1) { rep(i, 1, n) { if (s[i][1] == '0') continue; if (i != n) { f1({i + 1, 1}); } else { f4({n, 2}); } } } int xx = 1, yy = 1; if (n & 1) xx = 2; if (m & 1) yy = 2; // for (int i = 1; i <= n; ++i) { // for (int j = 1; j <= n; ++j) cout << s[i][j]; // puts(""); // } for (int i = xx; i <= n; i += 2) { for (int j = yy; j <= m; j += 2) { // cout << i << "---" << j << ' '; solve(i, j); } } cout << cnt << ' '; rep(i, 1, cnt) { printf("%d %d %d %d %d %d ", e[i].x1, e[i].y1, e[i].x2, e[i].y2, e[i].x3, e[i].y3); } // for (int i = 1; i <= n; ++i) { // for (int j = 1; j <= n; ++j) cout << s[i][j]; // puts(""); // } } // #ifndef ONLINE_JUDGE // system("pause"); // #endif }
D-Graph Subset Problem
不会写好难想不到
线段树搞一搞
维护一个区间最大、区间最小、区间和
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define ls u<<1 #define rs u<<1|1 #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const ll N = 2e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; ll n, q; ll a[N]; struct node { ll l, r; ll maxx, minn, lazy, sum; }tr[N << 2]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } void pushup(int u) { tr[u].maxx = max(tr[ls].maxx, tr[rs].maxx); tr[u].minn = min(tr[ls].minn, tr[rs].minn); tr[u].sum = tr[ls].sum + tr[rs].sum; } void change(int u, ll num) { tr[u].sum = (tr[u].r - tr[u].l + 1) * num; tr[u].minn = tr[u].maxx = num; tr[u].lazy = num; } void pushdown(int u) { change(ls, tr[u].lazy); change(rs, tr[u].lazy); tr[u].lazy = 0; } void build(int u, int l, int r) { tr[u].l = l; tr[u].r = r; tr[u].lazy = 0; if (l == r) { tr[u].minn = a[l]; tr[u].maxx = a[l]; tr[u].sum = a[l]; return ; } int mid = l + r >> 1; build(u << 1, l, mid); build(u << 1 | 1, mid + 1, r); pushup(u); } void upd(int u, int l, int r, ll val) { if (tr[u].minn > val) { return; } if (l <= tr[u].l && tr[u].r <= r) { if (tr[u].maxx <= val) { change(u, val); return; } } if (tr[u].lazy) pushdown(u); int mid = tr[u].l + tr[u].r >> 1; if (mid >= l) upd(ls, l, r, val); if (mid + 1 <= r) upd(rs, l, r, val); pushup(u); } int qry(int u, int l, int r, ll &val) { if (tr[u].minn > val) return 0; if (tr[u].l >= l && tr[u].r <= r) { if (tr[u].sum <= val) { val -= tr[u].sum; return tr[u].r - tr[u].l + 1; } } if (tr[u].lazy) pushdown(u); int mid = tr[u].l + tr[u].r >> 1; int ans1 = 0, ans2 = 0; if (mid >= l) ans1 = qry(ls, l, r, val); if (mid + 1 <= r) ans2 = qry(rs, l, r, val); return ans1 + ans2; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif n = read(); q = read(); rep(i, 1, n) a[i] = read(); build(1, 1, n); while (q--) { ll op, x, y; op = read(); x = read(); y = read(); if (op == 1) { upd(1, 1, x, y); } else if (op == 2) { cout << qry(1, x, n, y) << ' '; } } // #ifndef ONLINE_JUDGE // system("pause"); // #endif }