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;
}