2014-2015 ACM-ICPC, Asia Tokyo Regional Contest
2014-2015 ACM-ICPC, Asia Tokyo Regional Contest
| A | B | C | D | E | F | G | H | I | J | K |
| O | O | O | O | O | O |
签到
#include <bits/stdc++.h>
using namespace std; const int N = ;
int b[N], p[N], n, m, cnt[];
int temp[N]; int put(int opt) {
int cnt2[] = {};
int index = ;
for (int i = ; i <= m; i++) {
int num = p[i];
while (num--) {
temp[++index] = opt;
}
opt ^= ;
}
for (int i = ; i <= n; i++) cnt2[temp[i]]++;
if (cnt2[] != cnt[] || cnt2[] != cnt[]) return 1e9;
int j = ;
int res = ;
for (int i = ; i <= n; i++) {
if (b[i] == ) {
while (temp[j] != ) j++;
res += abs(i - j);
j++;
}
}
return res;
} int main() {
scanf("%d%d", &n, &m);
for (int i = ; i <= n; i++)
scanf("%d", &b[i]), cnt[b[i]]++;
for (int i = ; i <= m; i++)
scanf("%d", &p[i]);
printf("%d\n", min(put(), put()));
return ;
}
签到
#include <bits/stdc++.h>
using namespace std; char s[];
int res, st[];
int top; int main() {
scanf("%s%d", s + , &res);
int len = strlen(s + );
int ans1 = s[] - '';
for (int i = ; i <= len; i += ) {
if (s[i - ] == '+') {
ans1 += s[i] - '';
} else {
ans1 *= s[i] - '';
}
}
st[++top] = s[] - '';
for (int i = ; i <= len; i += ) {
if (s[i - ] == '*') {
st[top] *= s[i] - '';
} else {
st[++top] = s[i] - '';
}
}
int ans2 = ;
for (int i = ; i <= top; i++) ans2 += st[i];
if (ans1 == res && ans2 == res) puts("U");
else if (ans1 == res) puts("L");
else if (ans2 == res) puts("M");
else puts("I");
return ;
}
如果两个区间存在包含关系，那么肯定直接走到最大的 $d$ 再走到最小的 $c$ 最优。
如果两个区间交叉，那么之间应该是三段距离，分别设为 $a$, $b$, $c$
如果先走左区间再走右区间，那么走的路程为 $a + b + a + b + a + b + c + b + c + b + c= 3a + 5b + 3c$
如果先走右区间再走坐去间，那么走的路程为 $a + b + c + a + b + c + a + b + c = 3a + 3b + 3c$
所以就是把所有交叉的区间合并成一个大区间，然后每次到区间右端点再走回一遍左端点。
用并查集实现即可。
#include <bits/stdc++.h>
using namespace std; const int N = ;
int fa[N], mx[N], mn[N]; int getfa(int x) {
return x == fa[x] ? x : fa[x] = getfa(fa[x]);
} void unit(int x, int y) {
x = getfa(x), y = getfa(y);
if (x == y) return;
fa[x] = y;
mx[y] = max(mx[y], mx[x]);
mn[y] = min(mn[y], mn[x]);
} int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = ; i <= n; i++) fa[i] = mx[i] = mn[i] = i;
while (m--) {
int l, r;
scanf("%d%d", &l, &r);
for (int i = l; i < r; i++)
unit(i, i + );
}
int ans = n + ;
for (int i = ; i <= n; i++)
if (getfa(i) == i)
ans += (mx[i] - mn[i]) * ;
printf("%d\n", ans);
}
高中物理杀我！！！我高考考的最烂的就是物理了！！！
能不能跨过一个板只跟 $v_y$ 有关。枚举一下要弹多少次，然后二分一下 $v_y$，可以求出 $v_x$，得到的 $v_y$ 不一定使合速度最小，因为合速度是一个对勾函数，三分或者判断一下就行。
#include <bits/stdc++.h>
using namespace std; const int N = ;
const double eps = 1e-;
int d, n, b;
int p[N], h[N];
double loc[N]; int dcmp(double x) {
if (fabs(x) < eps) return ;
return x < ? - : ;
} bool check(int cnt, double vy) {
double per_time = vy * 2.0;
double total_time = per_time * cnt;
double vx = d / total_time;
for (int i = ; i <= cnt; i++) {
double l = loc[i - ], r = loc[i];
for (int j = ; j <= n; j++) {
if (dcmp(p[j] - l) < || dcmp(p[j] - r) > ) continue;
double cur_time = p[j] / vx;
int times = cur_time / per_time;
cur_time -= times * per_time;
if (dcmp(vy * cur_time - 0.5 * cur_time * cur_time - h[j]) < ) return ;
}
}
return ;
} double check0(int cnt, double vy) {
double per_time = vy * 2.0;
double total_time = per_time * cnt;
double vx = d / total_time;
return sqrt(vx * vx + vy * vy);
} int main() {
freopen("in.txt", "r", stdin);
scanf("%d%d%d", &d, &n, &b);
for (int i = ; i <= n; i++)
scanf("%d%d", p + i, h + i);
double ans = 2e4;
for (int i = ; i <= b + ; i++) {
bool flag = ;
if (d % i == ) {
int per = d / i;
for (int j = ; j <= n; j++)
if (p[j] % per == ) {
flag = ;
break;
}
}
if (flag) continue;
double per = 1.0 * d / i;
for (int j = ; j <= i; j++)
loc[j] = per * j;
double l = , r = 2e4;
for (int cnt = ; cnt < ; cnt++) {
double mid = (l + r) / 2.0;
if (check(i, mid)) r = mid;
else l = mid;
}
r = 2e4;
for (int cnt = ; cnt < ; cnt++) {
double ll = l + (r - l) / 3.0, rr = r - (r - l) / 3.0;
if (check0(i, ll) < check0(i, rr)) r = rr;
else l = ll;
}
ans = min(ans, check0(i, l));
}
printf("%.5f\n", ans);
return ;
}
这 $n$ 也出得太小了！！！本质就是求哪些边一定存在于所有最小生成树上。那么就是CF原题了。
#include <bits/stdc++.h>
using namespace std; const int N = 1e5 + ;
int n, m;
struct Edge {
int u, v, cost, id;
bool operator < (const Edge &rhs) const {
return cost < rhs.cost;
}
} edge[N], edge2[N]; struct E {
int v, ne, id;
} e[N];
int head[N], cnt, ans[N]; inline void add(int u, int v, int id) {
e[cnt] = (E){v, head[u], id};
head[u] = cnt++;
e[cnt] = (E){u, head[v], id};
head[v] = cnt++;
} int fa[N], dfn[N], low[N], tol;
int getfa(int x) { return x == fa[x] ? x : fa[x] = getfa(fa[x]); } void dfs(int u, int id) {
low[u] = dfn[u] = ++tol;
for (int i = head[u]; ~i; i = e[i].ne) {
int v = e[i].v;
if (!dfn[v]) {
dfs(v, i);
low[u] = min(low[u], low[v]);
if (low[v] > dfn[u]) ans[e[i].id] = ;
} else if (i != (id ^ )) {
low[u] = min(low[u], dfn[v]);
}
}
} inline void unite(int u, int v) {
u = getfa(u);
v = getfa(v);
if (u != v) fa[u] = v;
} int main() {
//freopen("in.txt", "r", stdin);
scanf("%d%d", &n, &m);
for (int i = ; i <= n; i++) fa[i] = i;
for (int i = ; i < m; i++) {
scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost);
edge[i].id = i;
}
memcpy(edge2, edge, sizeof(edge2));
sort(edge, edge + m);
for (int i = ; i < m; ) {
int j = i;
while (j + < m && edge[j + ].cost == edge[j].cost) j++;
cnt = tol = ;
for (int k = i; k <= j; k++) {
edge[k].u = getfa(edge[k].u);
edge[k].v = getfa(edge[k].v);
int u = edge[k].u, v = edge[k].v;
head[u] = head[v] = -;
dfn[u] = dfn[v] = low[u] = low[v] = ;
}
for (int k = i; k <= j; k++) {
int u = edge[k].u, v = edge[k].v, id = edge[k].id;
if (u == v) {
//printf("%d\n", k);
ans[id] = ;
continue;
}
ans[id] = ;
add(u, v, id);
}
for (int k = i; k <= j; k++)
if (!dfn[edge[k].u]) dfs(edge[k].u, -);
for (; i <= j; ++i)
unite(edge[i].u, edge[i].v);
}
int ans1 = ;
long long ans2 = ;
for (int i = ; i < m; i++)
if (ans[i] == ) ans1++, ans2 += edge2[i].cost;
printf("%d %lld\n", ans1, ans2);
return ;
}
这道题能写出来是因为大一上学期刚学C语言接触了括号序列判定是否合法的题。当时还没学栈，所以想了一个做法。把左括号看成 +$1$，右括号看成 -$1$，每个点表示当前的前缀和。
那么序列合法等价于，前缀和数组没有位置小于0，且最后一个数等于0。
对于这个题，在位置 $x$ 修改一个括号，就相当于区间 $[x, n]$ 加（减）2。
如果把一个左括号变成右括号，相当于区间 $-2$，那么只需要找当最左边的 $-1$位置，使它加 $2$ 就能使序列合法。
如果把一个右括号变成左括号，相当于区间 $+2$，那么找到第一个前缀和大于 $1$ 的位置，从他之后第一个 $1$ 变成 $-1$ 即可，这里我选择二分找那个位置。应该有1log的做法。
#include <bits/stdc++.h>
using namespace std; const int N = 3e5 + ;
int x[N], sum[N], n;
char s[N]; struct Seg {
#define lp p << 1
#define rp p << 1 | 1
int tree[N << ][];
void pushup(int p) {
tree[p][] = min(tree[lp][], tree[rp][]);
tree[p][] = min(tree[lp][], tree[rp][]);
}
void build(int p, int l, int r) {
if (l == r) {
if (l == || l == n) {
tree[p][] = tree[p][] = N;
return;
}
if (x[l] == ) tree[p][] = l, tree[p][] = N;
else tree[p][] = N, tree[p][] = l;
return;
}
int mid = l + r >> ;
build(lp, l, mid);
build(rp, mid + , r);
pushup(p);
}
void update(int p, int l, int r, int pos) {
if (l == r) {
swap(tree[p][], tree[p][]);
return;
}
int mid = l + r >> ;
if (pos <= mid) update(lp, l, mid, pos);
else update(rp, mid + , r, pos);
pushup(p);
}
int query(int p, int l, int r, int x, int y, int opt) {
if (x <= l && y >= r) return tree[p][opt];
int mid = l + r >> ;
int ans = N;
if (x <= mid) ans = min(ans, query(lp, l, mid, x, y, opt));
if (y > mid) ans = min(ans, query(rp, mid + , r, x, y, opt));
return ans;
}
} seg; struct Seg1 {
#define lp p << 1
#define rp p << 1 | 1
int tree[N << ], lazy[N << ];
void pushup(int p) {
tree[p] = min(tree[lp], tree[rp]);
}
void build(int p, int l, int r) {
if (l == r) {
tree[p] = sum[l];
return;
}
int mid = l + r >> ;
build(lp, l, mid);
build(rp, mid + , r);
pushup(p);
}
void tag(int p, int val) {
tree[p] += val;
lazy[p] += val;
}
void pushdown(int p) {
if (lazy[p]) {
tag(lp, lazy[p]);
tag(rp, lazy[p]);
lazy[p] = ;
}
}
void update(int p, int l, int r, int x, int y, int val) {
if (x <= l && y >= r) {
tree[p] += val;
lazy[p] += val;
return;
}
int mid = l + r >> ;
pushdown(p);
if (x <= mid) update(lp, l, mid, x, y, val);
if (y > mid) update(rp, mid + , r, x, y, val);
pushup(p);
}
int query(int p, int l, int r, int x, int y) {
if (x <= l && y >= r) return tree[p];
int mid = l + r >> ;
pushdown(p);
int ans = N;
if (x <= mid) ans = min(ans, query(lp, l, mid, x, y));
if (y > mid) ans = min(ans, query(rp, mid + , r, x, y));
return ans;
}
void print(int p, int l, int r){
if (l == r){
printf("%d ", tree[p]);
return;
}
int mid = l + r >> ;
pushdown(p);
print(lp, l, mid);
print(rp, mid + , r);
}
} seg1; int get(char ch) {
if (ch == '(') return ;
return -;
} int main() {
freopen("in.txt", "r", stdin);
int q;
scanf("%d%d", &n, &q);
scanf("%s", s + );
for (int i = ; i <= n; i++)
x[i] = get(s[i]), sum[i] = sum[i - ] + x[i];
seg.build(, , n);
seg1.build(, , n);
for (int u; q--; ) {
scanf("%d", &u);
if (u == || u == n) {
printf("%d\n", u);
continue;
}
seg.update(, , n, u);
if (x[u] == ) seg1.update(, , n, u, n, -);
else seg1.update(, , n, u, n, );
x[u] *= -;
int opt = (x[u] == ? : );
if (opt == ) {
int pos = seg.tree[][opt];
printf("%d\n", pos);
x[pos] *= -;
seg.update(, , n, pos);
seg1.update(, , n, pos, n, );
} else {
int l = , r = n;
int temp = ;
while (l <= r) {
int mid = l + r >> ;
if (seg1.query(, , n, mid, n) >= ) r = mid - , temp = mid;
else l = mid + ;
}
int pos = seg.query(, , n, temp, n, opt);
printf("%d\n", pos);
x[pos] *= -;
seg.update(, , n, pos);
seg1.update(, , n, pos, n, -);
}
}
return ;
}
还真有1log的
相当于查区间最后一个小于2的位置，那么先查右子树，如果右子树最小值小于2，就返回右，否则返回左。
#include <bits/stdc++.h>
using namespace std; const int N = 3e5 + ;
int x[N], sum[N], n;
char s[N]; struct Seg {
#define lp p << 1
#define rp p << 1 | 1
int tree[N << ][];
void pushup(int p) {
tree[p][] = min(tree[lp][], tree[rp][]);
tree[p][] = min(tree[lp][], tree[rp][]);
}
void build(int p, int l, int r) {
if (l == r) {
if (l == || l == n) {
tree[p][] = tree[p][] = N;
return;
}
if (x[l] == ) tree[p][] = l, tree[p][] = N;
else tree[p][] = N, tree[p][] = l;
return;
}
int mid = l + r >> ;
build(lp, l, mid);
build(rp, mid + , r);
pushup(p);
}
void update(int p, int l, int r, int pos) {
if (l == r) {
swap(tree[p][], tree[p][]);
return;
}
int mid = l + r >> ;
if (pos <= mid) update(lp, l, mid, pos);
else update(rp, mid + , r, pos);
pushup(p);
}
int query(int p, int l, int r, int x, int y, int opt) {
if (x <= l && y >= r) return tree[p][opt];
int mid = l + r >> ;
int ans = N;
if (x <= mid) ans = min(ans, query(lp, l, mid, x, y, opt));
if (y > mid) ans = min(ans, query(rp, mid + , r, x, y, opt));
return ans;
}
} seg; struct Seg1 {
#define lp p << 1
#define rp p << 1 | 1
int tree[N << ], lazy[N << ];
void pushup(int p) {
tree[p] = min(tree[lp], tree[rp]);
}
void build(int p, int l, int r) {
if (l == r) {
tree[p] = sum[l];
return;
}
int mid = l + r >> ;
build(lp, l, mid);
build(rp, mid + , r);
pushup(p);
}
void tag(int p, int val) {
tree[p] += val;
lazy[p] += val;
}
void pushdown(int p) {
if (lazy[p]) {
tag(lp, lazy[p]);
tag(rp, lazy[p]);
lazy[p] = ;
}
}
void update(int p, int l, int r, int x, int y, int val) {
if (x <= l && y >= r) {
tree[p] += val;
lazy[p] += val;
return;
}
int mid = l + r >> ;
pushdown(p);
if (x <= mid) update(lp, l, mid, x, y, val);
if (y > mid) update(rp, mid + , r, x, y, val);
pushup(p);
}
int query(int p, int l, int r) {
if (tree[p] > ) return l;
if (l == r) return l;
int mid = l + r >> ;
pushdown(p);
if (tree[rp] <= ) return query(rp, mid + , r);
return query(lp, l, mid);
}
void print(int p, int l, int r){
if (l == r){
printf("%d ", tree[p]);
return;
}
int mid = l + r >> ;
pushdown(p);
print(lp, l, mid);
print(rp, mid + , r);
}
} seg1; int get(char ch) {
if (ch == '(') return ;
return -;
} int main() {
freopen("in.txt", "r", stdin);
int q;
scanf("%d%d", &n, &q);
scanf("%s", s + );
for (int i = ; i <= n; i++)
x[i] = get(s[i]), sum[i] = sum[i - ] + x[i];
seg.build(, , n);
seg1.build(, , n);
for (int u; q--; ) {
scanf("%d", &u);
if (u == || u == n) {
printf("%d\n", u);
continue;
}
seg.update(, , n, u);
if (x[u] == ) seg1.update(, , n, u, n, -);
else seg1.update(, , n, u, n, );
x[u] *= -;
int opt = (x[u] == ? : );
if (opt == ) {
int pos = seg.tree[][opt];
printf("%d\n", pos);
x[pos] *= -;
seg.update(, , n, pos);
seg1.update(, , n, pos, n, );
} else {
int l = , r = n;
int temp = seg1.query(, , n) + ;
int pos = seg.query(, , n, temp, n, opt);
printf("%d\n", pos);
x[pos] *= -;
seg.update(, , n, pos);
seg1.update(, , n, pos, n, -);
}
}
return ;
}
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