Codeforces #264 (Div. 2) D. Gargari and Permutations
2014-09-07 21:58
585 查看
Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have foundk permutations. Each of them consists of numbers1, 2, ..., n
in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?
You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
Input
The first line contains two integers n andk(1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the nextk lines
contains integers1, 2, ..., n in some order — description of the current permutation.
Output
Print the length of the longest common subsequence.
Sample test(s)
Input
Output
Note
The answer for the first test sample is subsequence [1, 2, 3].
题意:求k个长度为n的最长公共子序列
思路1:保存每个数在各自串的位置,因为结果是第1个串中的某个可能,所以我们枚举第1个串的可能,然后检查如果一个以a[j]为结束的最长公共子序列成立的情况是,对于每个串的a[i]都在a[j]的前面,那么就有dp[j] = max(dp[j], dp[i]+1)
[cpp]
view plaincopyprint?
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1010;
int n, k;
int a[maxn][maxn], b[maxn][maxn], dp[maxn];
int check(int x,
int y) {
for (int i = 2; i <= k; i++)
if (b[i][x] > b[i][y])
return 0;
return 1;
}
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= k; i++)
for (int j = 1; j <= n; j++) {
scanf("%d", &a[i][j]);
b[i][a[i][j]] = j;
}
for (int i = 1; i <= n; i++)
dp[i] = 1;
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = i+1; j <= n; j++) {
if (check(a[1][i], a[1][j]))
dp[j] = max(dp[i]+1, dp[j]);
}
}
for (int i = 1; i <= n; i++)
ans = max(ans, dp[i]);
printf("%d\n", ans);
return 0;
}
思路2:如果一个数字i在每个串的位置都在j前面,那么i到j就有一条有向边,那么题目就转换为DAG求最长
[cpp]
view plaincopyprint?
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1010;
int num[10][maxn], vis[maxn];
int n, k;
vector<int> g[maxn];
int check(int x,
int y) {
for (int i = 0; i < k; i++)
if (num[i][x] >= num[i][y])
return 0;
return 1;
}
int dfs(int x) {
int ans = 0;
if (vis[x])
return vis[x];
int size = g[x].size();
for (int i = 0; i < size; i++)
ans = max(ans, dfs(g[x][i]));
return vis[x] = ans + 1;
}
int main() {
scanf("%d%d", &n, &k);
memset(vis, 0, sizeof(vis));
for (int i = 0; i <= n; i++)
g[i].clear();
int a;
for (int i = 0; i < k; i++)
for (int j = 1; j <= n; j++) {
scanf("%d", &a);
num[i][a] = j;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (check(i, j))
g[i].push_back(j);
int ans = 0;
for (int i = 1; i <= n; i++)
if (!vis[i])
ans = max(ans, dfs(i));
printf("%d\n", ans);
return 0;
}
in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?
You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
Input
The first line contains two integers n andk(1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the nextk lines
contains integers1, 2, ..., n in some order — description of the current permutation.
Output
Print the length of the longest common subsequence.
Sample test(s)
Input
4 3 1 4 2 3 4 1 2 3 1 2 4 3
Output
3
Note
The answer for the first test sample is subsequence [1, 2, 3].
题意:求k个长度为n的最长公共子序列
思路1:保存每个数在各自串的位置,因为结果是第1个串中的某个可能,所以我们枚举第1个串的可能,然后检查如果一个以a[j]为结束的最长公共子序列成立的情况是,对于每个串的a[i]都在a[j]的前面,那么就有dp[j] = max(dp[j], dp[i]+1)
[cpp]
view plaincopyprint?
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1010;
int n, k;
int a[maxn][maxn], b[maxn][maxn], dp[maxn];
int check(int x,
int y) {
for (int i = 2; i <= k; i++)
if (b[i][x] > b[i][y])
return 0;
return 1;
}
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= k; i++)
for (int j = 1; j <= n; j++) {
scanf("%d", &a[i][j]);
b[i][a[i][j]] = j;
}
for (int i = 1; i <= n; i++)
dp[i] = 1;
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = i+1; j <= n; j++) {
if (check(a[1][i], a[1][j]))
dp[j] = max(dp[i]+1, dp[j]);
}
}
for (int i = 1; i <= n; i++)
ans = max(ans, dp[i]);
printf("%d\n", ans);
return 0;
}
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int maxn = 1010; int n, k; int a[maxn][maxn], b[maxn][maxn], dp[maxn]; int check(int x, int y) { for (int i = 2; i <= k; i++) if (b[i][x] > b[i][y]) return 0; return 1; } int main() { scanf("%d%d", &n, &k); for (int i = 1; i <= k; i++) for (int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); b[i][a[i][j]] = j; } for (int i = 1; i <= n; i++) dp[i] = 1; int ans = 0; for (int i = 1; i <= n; i++) { for (int j = i+1; j <= n; j++) { if (check(a[1][i], a[1][j])) dp[j] = max(dp[i]+1, dp[j]); } } for (int i = 1; i <= n; i++) ans = max(ans, dp[i]); printf("%d\n", ans); return 0; }
思路2:如果一个数字i在每个串的位置都在j前面,那么i到j就有一条有向边,那么题目就转换为DAG求最长
[cpp]
view plaincopyprint?
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1010;
int num[10][maxn], vis[maxn];
int n, k;
vector<int> g[maxn];
int check(int x,
int y) {
for (int i = 0; i < k; i++)
if (num[i][x] >= num[i][y])
return 0;
return 1;
}
int dfs(int x) {
int ans = 0;
if (vis[x])
return vis[x];
int size = g[x].size();
for (int i = 0; i < size; i++)
ans = max(ans, dfs(g[x][i]));
return vis[x] = ans + 1;
}
int main() {
scanf("%d%d", &n, &k);
memset(vis, 0, sizeof(vis));
for (int i = 0; i <= n; i++)
g[i].clear();
int a;
for (int i = 0; i < k; i++)
for (int j = 1; j <= n; j++) {
scanf("%d", &a);
num[i][a] = j;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (check(i, j))
g[i].push_back(j);
int ans = 0;
for (int i = 1; i <= n; i++)
if (!vis[i])
ans = max(ans, dfs(i));
printf("%d\n", ans);
return 0;
}
相关文章推荐
- Codeforces #264(div 2)D.Gargari and Permutations
- Codeforces #264 (Div. 2) D. Gargari and Permutations(DAG求最长路)
- Codeforces #264 (Div. 2) D. Gargari and Permutations(动态规划:简单)
- Codeforces 264 (Div. 2) D. Gargari and Permutations
- Codeforces #264 (Div. 2) D. Gargari and Permutations
- Codeforces #264 div.2 D. Gargari and Permutations
- Codeforces Round #264 (Div. 2) D. Gargari and Permutations 多序列LIS+dp好题
- Codeforces #264 (Div. 2) C. Gargari and Bishops
- Codeforces #264 (Div. 2) D. Gargari and Permutations
- Codeforces Round #264 (Div. 2) D. Gargari and Permutations 多序列LIS+dp好题
- codeforce 263 div2D Gargari and Permutations
- Codeforces Round #264 (Div. 2) C. Gargari and Bishops
- 【CODEFORCES】 D. Gargari and Permutations
- Codeforces Round #285 (Div.1 B & Div.2 D) Misha and Permutations Summation --二分+树状数组
- Codeforces Round #285 (Div. 2) D. Misha and Permutations Summation 康托展开 树状数组+二分
- Codeforces Round 264(div2) C. Gargari and Bishops
- COdeforces Round #264 (Div. 2) C. Gargari and Bishops
- Codeforces Round #264 (Div. 2)-C. Gargari and Bishops
- 【【henuacm2016级暑期训练】动态规划专题 I】Gargari and Permutations
- Codeforces Round #285 Div.1 B Misha and Permutations Summation