比赛-xxxxxyt学姐的训练赛 (25 Aug, 2018)
NKOJ 上可以搜索到题目。
A. Candies##
BFS 一下就可以了。注意这题不满足二分性质。
#include <cstdio>
#include <queue>
using namespace std;
struct node {
int x, y;
node(int x = 0, int y = 0):
x(x), y(y) { }
};
queue<node> Q;
int A[105][105], V, N, M, Cnt;
int nxt[4][2] = {0,1,0,-1,1,0,-1,0};
bool mk[105][105], mk2[105][105];
void check(int x, int y)
{
if (1 <= x && x <= N && 1 <= y && y <= M && A[x][y] >= V && !mk[x][y])
Q.push(node(x, y)), mk[x][y] = 1, ++Cnt;
return;
}
bool test()
{
Cnt = 0;
for (int i = 1; i <= N; ++i)
for (int j = 1; j <= M; ++j)
mk[i][j] = 0;
for (int i = 1; i <= N; ++i) {
check(i, 1);
check(i, M);
}
for (int j = 2; j < M; ++j) {
check(1, j);
check(N, j);
}
while (!Q.empty()) {
node p = Q.front();
Q.pop();
for (int i = 0; i < 4; ++i)
check(p.x + nxt[i][0], p.y + nxt[i][1]);
}
if (Cnt == N * M) return 0;
bool flag = 0;
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= M; ++j) {
if (!mk[i][j]) {
Q.push(node(i, j));
flag = 1;
break;
}
}
if (flag) break;
}
for (int i = 1; i <= N; ++i)
for (int j = 1; j <= M; ++j)
mk2[i][j] = 0;
int tot = 0;
while (!Q.empty()) {
node p = Q.front();
Q.pop();
for (int i = 0; i < 4; ++i) {
int x = p.x + nxt[i][0], y = p.y + nxt[i][1];
if (1 <= x && x <= N && 1 <= y && y <= M && !mk[x][y] && !mk2[x][y])
Q.push(node(x, y)), mk2[x][y] = 1, ++tot;
}
}
return tot + Cnt < N * M;
}
int main()
{
scanf("%d%d", &N, &M);
for (int i = 1; i <= N; ++i)
for (int j = 1; j <= M; ++j)
scanf("%d", &A[i][j]);
for (V = 100; V >= 1; --V) {
if (test()) {
printf("%d\n", V);
return 0;
}
}
printf("-1\n");
return 0;
}
B. Cards##
BZOJ3262 陌上花开 弱化版，没有相同元素。三位偏序问题（第一维排序，第二维分治，第三维数据结构）。同时这题非常良心，不同数据采用不同算法，即暴力 dp 或者 topsort 可过 60 分的数据， \(O(n^2)\) ；剩下的 40 分是一个二维偏序，用 cdq 分治，或者对第一维排序后 \(O(n \cdot log\ n)\) 求最长不降子序列即可。
60 + 40 分的代码
#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int _N = 1020000;
int ind[_N], len[_N], N;
vector<int> G[_N];
queue<int> Q;
struct data {
int a, b, c;
bool operator < (const data &tmp)
const
{
return a <= tmp.a && b <= tmp.b && c <= tmp.c;
}
} A[_N];
bool cmp(data a, data b)
{
return a.a < b.a || a.a == b.a && a.b < b.b;
}
void fun1()
{
int ans;
sort(A + 1, A + 1 + N, cmp);
len[ans = 1] = A[1].b;
for (int i = 2; i <= N; ++i) {
if (A[i].b >= len[ans]) {
len[++ans] = A[i].b;
} else {
int t = upper_bound(len + 1, len + 1 + ans, A[i].b) - len;
len[t] = A[i].b;
}
}
printf("%d\n", ans);
return;
}
int main()
{
scanf("%d", &N);
for (int i = 1; i <= N; ++i)
scanf("%d%d%d", &A[i].a, &A[i].b, &A[i].c);
if (N >= 1055) {
fun1();
return 0;
}
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= N; ++j) {
if (i != j && A[i] < A[j])
G[i].push_back(j), ++ind[j];
}
}
for (int i = 1; i <= N; ++i)
if (!ind[i]) Q.push(i), len[i] = 1;
while (!Q.empty()) {
int p = Q.front();
Q.pop();
for (int i = G[p].size() - 1; i >= 0; --i) {
int v = G[p][i];
if (!--ind[v]) Q.push(v), len[v] = len[p] + 1;
}
}
int ans = 0;
for (int i = 1; i <= N; ++i)
ans = max(ans, len[i]);
printf("%d\n", ans);
return 0;
}
C. Pets##
分析一下可以发现两部分的图中均不能有环， topsort 判一下是否有环即可。然后问题转化成要把第二部分链中的节点插入到第一部分的链中（ topsort 之后肯定保证是一条链）。 \(f_{i,j}\) 表示考虑把第二部分链第 \(j\) 个点插入到第一部分链第 \(i\) 和 \(i+1\) 个点中间，则：
\[f_{i,j} = max(f_{i-1,j},f_{i,j-1}+[t])
\]
事件 \(t\) 是指第二部分第 \(j\) 个点可以合法插入，为真条件是第一部分链 \([1,i]\) 这些点均有到 \(j\) 的边，且 \([i+1,m]\) 这些点均没有到 \(j\) 的边（这也意味着 \(j\) 到这些点有边）。这个可以 \(O(n^2)\) 预处理。答案是 \(f_{m,n-m}\) 。
#include <stdio.h>
#include <vector>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int _N = 1200;
int top[2][_N], f[_N][_N], ind[_N], pos[_N], N, M;
bool mk[_N], G[_N][_N];
queue<int> Q;
int topsort(bool t)
{
for (int i = 1; i <= N; ++i) ind[i] = 0;
for (int i = 1; i <= N; ++i)
for (int j = 1; j <= N; ++j)
if (mk[i] ^ t && mk[j] ^ t && G[i][j])
++ind[j];
for (int i = 1; i <= N; ++i)
if (mk[i] ^ t && !ind[i]) Q.push(i);
int cnt = 0;
while (!Q.empty()) {
int p = Q.front();
Q.pop(), top[t][++cnt] = p;
for (int i = 1; i <= N; ++i)
if (mk[i] ^ t && G[p][i] && !--ind[i])
Q.push(i);
}
return cnt;
}
int dfs(int i, int j)
{
if (~f[i][j]) return f[i][j];
if (!j) return f[i][j] = 0;
if (!i) return f[i][j] = dfs(i, j - 1) + (pos[j] == i);
return f[i][j] = max(dfs(i - 1, j), dfs(i, j - 1) + (pos[j] == i));
}
int main()
{
scanf("%d%d", &N, &M);
for (int i = 1; i <= N; ++i)
for (int j = 1; j <= N; ++j)
scanf("%d", &G[i][j]);
for (int t, i = 1; i <= M; ++i)
scanf("%d", &t), mk[t] = 1;
if (topsort(1) != N - M || topsort(0) != M) {
printf("NO\n");
return 0;
}
for (int i = 1; i <= N - M; ++i) {
int j, t;
for (j = 1; j <= M; ++j)
if (!G[top[0][j]][top[1][i]]) break;
for (t = j; j <= M; ++j)
if (G[top[0][j]][top[1][i]]) break;
pos[i] = j == M + 1 ? t - 1 : -1;
}
memset(f, -1, sizeof f);
printf("YES %d\n", dfs(M, N - M));
return 0;
}
