The 2014 ACM-ICPC Asia Mudanjiang Regional Contest
没去现场。做的网络同步赛。感觉还能够，搞了6题
A：这是签到题，对于A堆除掉。假设没剩余在减一。B堆直接除掉 + 1就能够了
B：二分贪心，二分长度。然后会发现本质上是在树上最长链上找两点，那么有二分出来的长度了，就从两端分别往里移动那么长，那两个位置就是放置位置。然后在推断一下就能够了
D：概率DP。首先知道放一个棋子。能够等价移动到右上角区域，那么就能够dp[x][y][k]，表示已经完毕x行y列。已经放了k个棋子，那么棋盘被划分成4个区域。利用这个区域能够算概率，然后去转移就可以
H：字符串处理，先hash掉，记录每一个hash值相应的l，r区间。离线处理完每次询问输出就能够了，注意细节
I：签到题，注意0的时候是等于0就能够了
K：贪心，假设数字个数不够。在前面补是最优的。然后从左往右扫描一遍，每次遇到1个星号要扣掉两个数字，假设没有，这须要交换操作。交换肯定和最后面的数字交换是最优的，最后还要考虑一种特殊情况。就是没有交换，而且最后还有数字，这时候要加入一个星号
代码：
A：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; const int N = 55;
int t, n, m; int a, b, suma, sumb; int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
suma = sumb = 0;
for (int i = 0; i < n - 1; i++) {
scanf("%d", &a);
suma += a;
}
for (int i = 0; i < m; i++) {
scanf("%d", &b);
sumb += b;
}
int l, r;
l = suma / (n - 1) - (suma % (n - 1) == 0);
r = sumb / m + 1;
if (l > r) swap(l, r);
printf("%d %d\n", l, r);
}
return 0;
}
B：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; const int N = 200005; struct Edge {
int u, v;
Edge() {}
Edge(int u, int v) {
this->u = u;
this->v = v;
}
} E[N * 2]; int first[N], next[N * 2], en; void add_Edge(int u, int v) {
E[en] = Edge(u, v);
next[en] = first[u];
first[u] = en++;
} const int INF = 0x3f3f3f3f;
int t, n, vis[N], f[N], u1, u2, Q[N * 2]; int bfs(int root) {
int head = 0, rear = 0;
vis[root] = 0;
Q[rear++] = root;
f[root] = 0;
int u;
while (head < rear) {
u = Q[head++];
for (int i = first[u]; i + 1; i = next[i]) {
Edge e = E[i];
int v = e.v;
if (vis[v] > vis[u] + 1) {
vis[v] = vis[u] + 1;
f[v] = u;
Q[rear++] = v;
}
}
}
return u;
} bool judge(int len) {
memset(vis, INF, sizeof(vis));
u1 = bfs(1);
for (int i = 0; i < len; i++) {
if (f[u1] == 0) {
u2 = u1 % n + 1;
return true;
}
u1 = f[u1];
}
memset(vis, INF, sizeof(vis));
u2 = bfs(u1);
for (int i = 0; i < len; i++) {
if (f[u2] == 0) return true;
u2 = f[u2];
}
bfs(u2);
for (int i = 1; i <= n; i++)
if (vis[i] > len) return false;
return true;
} int main() {
scanf("%d", &t);
while (t--) {
en = 0;
memset(first, -1, sizeof(first));
scanf("%d", &n);
int u, v;
for (int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
add_Edge(u, v);
add_Edge(v, u);
}
if (n == 2) {
printf("0 1 2\n");
continue;
}
int l = 0, r = n;
while (l < r) {
int mid = (l + r) / 2;
if (judge(mid)) r = mid;
else l = mid + 1;
}
judge(l);
if (u1 == u2) u2 = u2 % n + 1;
printf("%d %d %d\n", l, u1, u2);
}
return 0;
}
D：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; const int N = 55;
int t, n, m;
double dp[N][N][N * N]; int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
for (int k = max(i, j); k <= i * j; k++) {
double p1 = (i * j - (k - 1)) * 1.0;
double p2 = (n - i + 1) * j;
double p3 = i * (m - j + 1);
double p4 = (n - i + 1) * (m - j + 1);
dp[i][j][k] = dp[i][j][k - 1] * p1 + dp[i - 1][j][k - 1] * p2 + dp[i][j - 1][k - 1] * p3 + dp[i - 1][j - 1][k - 1] * p4;
dp[i][j][k] /= (n * m - (k - 1));
}
}
}
double ans = 0;
for (int i = max(n, m); i <= n * m; i++)
ans += dp[n][m][i] * i - dp[n][m][i - 1] * i;
printf("%.12lf\n", ans);
}
return 0;
}
H：
#include <cstdio>
#include <cstring>
#include <map>
using namespace std; #define MP(a,b) make_pair(a,b) typedef unsigned long long ull;
const ull x = 123;
const int N = 200005;
typedef pair<int, int> pii; char str[N], q[N];
map<ull, pii> g; int t, u, k, lens; int build(ull st) {
int flag = 1;
while (u < lens) {
if (str[u] == '{') {
flag = 0;
u++;
}
if (str[u] == '}') {
u++;
return u;
}
ull nt = st;
while (str[u] != ':' && str[u] != ',' && str[u] != '}')
nt = nt * x + str[u++];
if (str[u] == ':') {
int l = ++u;
int r = build(nt * x + '.');
g[nt] = MP(l, r);
}
if (str[u] == '}') {
if (!flag)
u++;
return u;
}
if (str[u] == ',') {
if (flag) return u;
u++;
}
}
} int main() {
scanf("%d", &t);
while (t--) {
u = 0;
g.clear();
scanf("%s", str);
lens = strlen(str);
build(0);
scanf("%d", &k);
while (k--) {
scanf("%s", q);
ull tmp = 0;
int len = strlen(q);
for (int i = 0; i < len; i++)
tmp = tmp * x + q[i];
if (!g.count(tmp)) {
printf("Error!\n");
continue;
}
int l = g[tmp].first, r = g[tmp].second;
for (int i = l; i < r; i++)
printf("%c", str[i]);
printf("\n");
}
}
return 0;
}
I：
#include <cstdio>
#include <cstring>
#include <cmath> int t, n;
char str[10]; double solve(int x) {
if (x == 0) return 0;
double sb = x * 1.0 / 100;
if (str[0] == 'b') return sb * log2(sb);
if (str[0] == 'n') return sb * log(sb);
if (str[0] == 'd') return sb * log10(sb);
} int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%s", &n, str);
double ans = 0;
int x;
for (int i = 0; i < n; i++) {
scanf("%d", &x);
ans += solve(x);
}
printf("%.12lf\n", -ans);
}
return 0;
}
K：
#include <cstdio>
#include <cstring>
#include <map>
using namespace std; #define MP(a,b) make_pair(a,b) typedef unsigned long long ull;
const ull x = 123;
const int N = 200005;
typedef pair<int, int> pii; char str[N], q[N];
map<ull, pii> g; int t, u, k, lens; int build(ull st) {
int flag = 1;
while (u < lens) {
if (str[u] == '{') {
flag = 0;
u++;
}
if (str[u] == '}') {
u++;
return u;
}
ull nt = st;
while (str[u] != ':' && str[u] != ',' && str[u] != '}')
nt = nt * x + str[u++];
if (str[u] == ':') {
int l = ++u;
int r = build(nt * x + '.');
g[nt] = MP(l, r);
}
if (str[u] == '}') {
if (!flag)
u++;
return u;
}
if (str[u] == ',') {
if (flag) return u;
u++;
}
}
} int main() {
scanf("%d", &t);
while (t--) {
u = 0;
g.clear();
scanf("%s", str);
lens = strlen(str);
build(0);
scanf("%d", &k);
while (k--) {
scanf("%s", q);
ull tmp = 0;
int len = strlen(q);
for (int i = 0; i < len; i++)
tmp = tmp * x + q[i];
if (!g.count(tmp)) {
printf("Error!\n");
continue;
}
int l = g[tmp].first, r = g[tmp].second;
for (int i = l; i < r; i++)
printf("%c", str[i]);
printf("\n");
}
}
return 0;
}