题目：https://www.nowcoder.com/question/next?pid=16516564&qid=362292&tid=24192838
解法一：dp,过程有点繁琐，主要是集合的有关运算
#include
using namespace std;
class Feature{
public:
int x;
int y;
Feature():x(0),y(0){
}
Feature(int _x,int _y):x(_x),y(_y){
}
/*当用set存储自定义类型数据时
需要重载<操作符，用于指定数据在set内的排序准则
set的插入过程:
其实，set容器在判定已有元素a和新插入元素b是否相等时，是这么做的：
1）将a作为左操作数，b作为有操作数，调用比较函数，并返回比较值
2）将b作为左操作数，a作为有操作数，再调用一次比较函数，并返回比较值
如果1、2两步的返回值都是false，则认为a、b是相等的，则b不会被插入set容器中
如果1、2两步的返回值都是true，则可能发生未知行为
因此，记住一个准则：永远让比较函数对相同元素返回false
*/
bool operator < (const Feature &f) const{
if(this->x == f.x)
return this->y < f.y;
return this->x < f.x;
}
};
class Frame{
public:
int num;
vector
};
void HasCommon(vector
ret.clear();
if(f1.size() == 0 || f2.size() == 0)
return;
set
for(auto i:f1)
set1.insert(i);
set
for(auto i:f2)
set2.insert(i);
set
set_intersection(set1.begin(),set1.end(),set2.begin(),set2.end(),inserter(common,common.begin()));
for(auto f:common)
ret.push_back(f);
}
int func(vector &frames){
int sz = frames.size();
/*dp[i]表示以frames[i]结尾的最长连续特征*/
vector
for(auto i = 0;i < sz;++i)
dp[i] = frames[i].num != 0 ? 1 : 0;
int ans = dp[0];
for(auto i = 1;i if(dp[i] == 0) continue; int j = i - 1; vector HasCommon(frames[j].features,frames[i].features,common); while(j >= 0 && common.size() > 0){ --j; vector copy(common.begin(),common.end(),tmp.begin()); HasCommon(tmp,frames[j].features,common); } dp[i] = i - j; ans = max(ans,dp[i]); } return ans; } int main(){ int N,M; cin >> N; for(auto i = 0;i < N;++i){ cin >> M; vector frames(M); for(auto j = 0;j < M;++j){ Frame tmp; cin >> tmp.num; tmp.features.resize(tmp.num); for(auto k = 0;k < tmp.num;++k){ cin >> tmp.features[k].x; cin >> tmp.features[k].y; } frames[j] = tmp; } cout< } return 0; } 解法二：哈希，把每个二维特征(x,y)映射成一个数字 #include using namespace std; class Frame{ public: int num; vector }; int func(vector &frames){ int sz = frames.size(); /*记录到上一帧为止最大连续特征数量*/ map /*这里有个假设：对于两个特征(x1,y1),(x2,y2) A*x1 + y1 = A*x2+y2当且仅当 x1 = x2,y1=y2 */ const uint64_t A = 10000481; int cur = 0,ret = 1; uint64_t key; for(auto i = 0;i < sz;++i){ cur = frames[i].num; if(cur == 0) continue; map for(auto j = 0;j < cur;++j){ key = A * frames[i].features[j].first + frames[i].features[j].second; if(cnt.find(key) != cnt.end()){ tmp[key] = cnt[key] + 1; ret = max(ret,tmp[key]); } else tmp[key] = 1; } cnt = tmp; } return ret; } int main(){ int N,M; cin >> N; for(auto i = 0;i < N;++i){ cin >> M; vector frames(M); for(auto j = 0;j < M;++j){ Frame tmp; cin >> tmp.num; tmp.features.resize(tmp.num); for(auto k = 0;k < tmp.num;++k){ cin >> tmp.features[k].first; cin >> tmp.features[k].second; } frames[j] = tmp; } cout< } return 0; } 解法三：https://blog.csdn.net/wwxy1995/article/details/89463465 #include using namespace std; typedef pair class Frame{ public: int num; vector }; int func(vector &frames){ int sz = frames.size(); /*记录每一个特征的上一次帧索引*/ map /*记录每一个特征最大连续次数*/ map for(auto i = 0;i lastIndex[frames[0].features[i]] = 0; cnt[frames[0].features[i]] = 1; } int ret = 1; for(auto i = 1;i < sz;++i){ for(auto j = 0;j if(lastIndex[frames[i].features[j]] == i-1) cnt[frames[i].features[j]] += 1; else cnt[frames[i].features[j]] = 1; lastIndex[frames[i].features[j]] = i; ret = max(ret,cnt[frames[i].features[j]]); } } return ret; } int main(){ int N,M; cin >> N; for(auto i = 0;i < N;++i){ cin >> M; vector frames(M); for(auto j = 0;j < M;++j){ Frame tmp; cin >> tmp.num; tmp.features.resize(tmp.num); for(auto k = 0;k < tmp.num;++k){ cin >> tmp.features[k].first; cin >> tmp.features[k].second; } frames[j] = tmp; } cout< } return 0; }