Full_of_Boys训练3总结
题目来源: 2016-2017 ACM-ICPC Pacific Northwest Regional Contest
E.Enclosure
先计算出内外两个凸包,枚举大凸包上的点,在小凸包上找到两个切点。计算面积时,就相当于删掉几条原先的边,加上一个新的三角形。同时,可以注意到,如果我们按照顺时针枚举大凸包上的点,那两个切点也只可能朝顺时针方向移动,这样的话,就可以在线性时间内计算出新的切点了。以后补代码。
G.Maximum Islands
贪心方法:把原本的L四个方向的C改成W,然后剩余的C,可以运用最小割的思想,用有效点数减最小割,就是最大的答案。思想来自骑士共存。二分图用的匈牙利算法。
#include <bits/stdc++.h> #define rg register #define pb(x) push_back(x) typedef long long ll; inline int read() { char c=getchar();int 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; } inline write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); } using namespace std; int n, m, col[45][45], dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}, vis[44][44]; char s[45][45]; vector<int> v[2]; struct node{ int x, y; node(){} node(int a,int b) {x = a; y = b;} }; inline int inb(int x, int y) { if(x < 1 || x > n || y < 1 || y > m) return 0; return 1; } inline void bfs1(int sx, int sy) { queue<node> q; q.push(node(sx, sy)); col[sx][sy] = 0; v[0].pb((sx-1)*m+sy); while(!q.empty()) { node t = q.front(); q.pop(); for(rg int i = 0; i < 4; ++i){ int tx = t.x + dx[i], ty = t.y + dy[i]; if(inb(tx, ty) && s[tx][ty] == 'C' && col[tx][ty] == -1) { col[tx][ty] = col[t.x][t.y]^1; v[col[tx][ty]].pb((tx-1)*m+ty); q.push(node(tx, ty)); } } } } struct edge{int e, nxt;}E[55*55*4]; int h[55*55], cc; inline void add(int u, int v) { E[cc].e = v; E[cc]. nxt = h[u]; h[u] = cc; ++cc; } inline void build() { memset(h, -1, sizeof(h)); cc = 0; for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j <= m; ++j) if(col[i][j] == 0){ for(rg int k = 0; k < 4; ++k) { int tx = i + dx[k], ty = j + dy[k]; if(inb(tx, ty) && col[tx][ty] == 1) add((i-1)*m+j, (tx-1)*m+ty); } } } inline int bfs2(int sx, int sy, char c) { int ans = 1; vis[sx][sy] = 1; queue<node> q; q.push(node(sx, sy)); while(!q.empty()) { node t = q.front(); q.pop(); for(rg int i = 0; i < 4; ++i) { int tx = t.x + dx[i], ty = t.y + dy[i]; if(inb(tx, ty) && s[tx][ty] == c && !vis[tx][ty]) vis[tx][ty]=1,++ans,q.push(node(tx,ty)); } } return ans; } int used[55*55], lk[55*55]; inline int dfs(int u) { for(rg int i = h[u]; ~i; i = E[i].nxt) if(!used[E[i].e]) { used[E[i].e] = 1; if(lk[E[i].e] == -1 || dfs(lk[E[i].e])) { lk[E[i].e] = u; return 1; } } return 0; } inline int hungray() { int ans = 0; memset(lk, -1, sizeof(lk)); for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j <= m; ++j) if(col[i][j] == 0){ memset(used, 0, sizeof(used)); if(dfs((i-1)*m+j)) ++ans; } return ans; } inline int solve() { int ans = 0, t = 0; for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j<= m; ++j) if(s[i][j] == 'L' && !vis[i][j]) bfs2(i, j, 'L'), ++ans; memset(vis, 0 , sizeof(vis)); for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j<= m; ++j) if(s[i][j] == 'C' && !vis[i][j]) t += bfs2(i, j, 'C'); return ans + t - hungray(); } int main() { n=read(),m=read(); memset(col, -1, sizeof(col)); for(rg int i = 1; i <= n; ++i) scanf(" %s",s[i]+1); for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j <= m; ++j) if(s[i][j]=='L') { for(rg int k = 0; k < 4; ++k) if(inb(i+dx[k],j+dy[k])&&s[i+dx[k]][j+dy[k]]=='C') { s[i+dx[k]][j+dy[k]] = 'W'; } } for(rg int i = 1; i <= n; ++i) for(rg int j = 1; j <= m; ++j) if(s[i][j] == 'C'&&col[i][j] == -1) { bfs1(i, j); } build(); write(solve()); return 0; }
J.Shopping
每次找到区间内最左边的小于x的数,然后%一下它,重复以上操作就行了。所以只需要实现一个区间询问最左边的小于x的值就可以了。可以证明每次操作最多log次
解法1:分块。块外暴力,块内提前排好序二分。写挫了莫名t。
解法2:线段树。维护一下区间最小值,显然如果左边的最小值小于等于x那就朝左边递归,否则右边,便可以完成这个操作。
解法3:st表。又不用修改。st表干掉一个log
code:
解法1:挫了不贴了。
解法2:
#include <cstdio> #include <algorithm> #define pb(x) push_back(x) #define rg register const int maxn = 2e5 + 100; typedef unsigned long long ll; inline int readint(){ char c=getchar();int 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; } inline ll readll(){ 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; } inline void write(ll x){ if(x>=10LL) write(x/10LL); putchar('0'+x%10LL); } using namespace std; int n, q; ll a[maxn]; struct seg{int l, r; ll x;} tr[maxn << 2]; inline void build(int p, int l, int r) { tr[p].l = l; tr[p].r = r; if(l == r){ tr[p].x = a[l]; return; } int mid = (l + r) >> 1; build(p<<1, l, mid); build(p<<1|1, mid+1, r); tr[p].x = min(tr[p<<1].x, tr[p<<1|1].x); } inline ll ask_mn(int p, int l, int r) { if(tr[p].l == l && tr[p].r == r) return tr[p].x; int mid = (tr[p].l + tr[p].r) >> 1; if(r <= mid) return ask_mn(p<<1, l, r); else if(l > mid) return ask_mn(p<<1|1, l, r); else return min(ask_mn(p<<1, l, mid), ask_mn(p<<1|1, mid+1, r)); } inline ll fd(ll x, int L, int R) { if(L == R) return L; int mid =(L + R) >> 1; if(ask_mn(1, L, mid) <= x) return fd(x, L, mid); else return fd(x, mid+1, R); } inline ll solve(ll x, int L, int R){ while(ask_mn(1, L, R) <= x) x %= a[fd(x, L, R)]; return x; } int main() { n = readint(), q = readint(); for(int i = 1; i <= n; ++i) a[i] = readll(); build(1, 1, n); while(q--) { ll x; int L, R; x = readll(), L = readint(), R = readint(); write(solve(x, L, R)); puts(""); } return 0; }
解法3:
#include <cstdio> #include <algorithm> #include <iostream> #define pb(x) push_back(x) #define rg register const int maxn = 2e5 + 100; typedef unsigned long long ll; inline int readint(){ char c=getchar();int 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; } inline ll readll(){ 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; } inline void write(ll x){ if(x>=10LL) write(x/10LL); putchar('0'+x%10LL); } using namespace std; int n, q; ll a[maxn], st[maxn][25]; inline void RMQ_init(){ for(int i = 0; i <= n; ++i) st[i][0] = a[i]; for(int j = 1; (1<<j) <= n; ++j){ for(int i = 1; i+(1<<j)-1 <= n; ++i){ st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]); } } } inline ll RMQ(int u, int v){ int k = (int)(log(v-u+1.0)/log(2.0)); return min(st[u][k], st[v-(1<<k)+1][k]); } inline ll fd(ll x, int L, int R) { if(L == R) return L; int mid =(L + R) >> 1; if(RMQ(L, mid) <= x) return fd(x, L, mid); else return fd(x, mid+1, R); } inline ll solve(ll x, int L, int R){ while(RMQ(L, R) <= x) x %= a[fd(x, L, R)]; return x; } int main() { n = readint(), q = readint(); for(int i = 1; i <= n; ++i) a[i] = readll(); RMQ_init(); while(q--) { ll x; int L, R; x = readll(), L = readint(), R = readint(); write(solve(x, L, R)); puts(""); } return 0; }