uva 10635 - Prince and Princess
2013-10-16 13:08
381 查看
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <map> #include <algorithm> #define maxn 250*250 + 10 #define ll long long #define INF 1000000000 #define FOR(i, a, b) for(int i = a; i < b; ++i) using namespace std; int main() { int kase = 0, t; scanf("%d", &t); while(t--) { int n, p, q; int a[maxn], b[maxn], g[maxn], len_b = 0, num; scanf("%d%d%d", &n, &p, &q); memset(a, 0, sizeof(a)); FOR(i, 1, p+2) {scanf("%d", &num); a[num] = i;} FOR(i, 1, q+2) { scanf("%d", &num); if(a[num]) b[len_b++] = a[num]; } FOR(i, 0, len_b+1) g[i] = INF; int ans = 0; int ss[maxn]; FOR(i, 0, len_b){ int k = lower_bound(g+1, g+len_b+1, b[i]) - g; //ss[i] = k; g[k] = b[i]; ans = max(ans, k); } //FOR(i, 0, len_b) printf("%d ", ss[i]); printf("\n"); printf("Case %d: %d\n", ++kase, ans); } return 0; }
相关文章推荐
- UVA 10635 Prince and Princess
- UVA 10635 Prince and Princess【LCS 问题转换为 LIS】
- UVA 10635 Prince and Princess
- 一中OJ #1434 序列的LCS [UVa 10635 -> Prince and Princess] | 动态规划 LCS映射转换LIS | 解题报告
- UVA 10635 Prince and Princess (dp + LCS)
- uva 10635 - Prince and Princess(LCS)
- UVA - 10635 —— Prince and Princess
- UVA 10635 Prince and Princess
- UVA 10635 Prince and Princess【LCS 问题转换为 LIS】
- uva 10635 Prince and Princess(稀疏LCS)
- UVA - 10635 Prince and Princess
- UVa 10635 (LIS+二分) Prince and Princess
- 【UVA 10635】【LCS转化成LIS】 Prince and Princess
- UVA - 10635 Prince and Princess
- uva 10635 Prince and Princess(LCS成问题LIS问题O(nlogn))
- Prince and Princess UVA - 10635(LCS转LIS)
- UVa 10635 Prince and Princess - 动态规划
- Uva 10635 - Prince and Princess(LIS)
- UVA - 10635 Prince and Princess (二分法 求最长上升子序列)
- uva 10635 - Prince and Princess