ZOJ-3721

这题有点意思，勉强可以归于DP吧，跑步时获得的思路，哈哈，写完1A了，思路是对的，略爽，归纳一下就是找出最大的互不相交的线段组，比如[1,2], [2,3]就是两组，而[1,3],[1,2],[2,4],[2,3]都可以归为同一组，先排序预处理一下，然后有O(n)算法可求组，附上两段代码，第一段是初版，内存使用较大，第二段优化了下内存，但时间没怎么变

#include<cstdio>
#include<vector>
#include<cstdlib>

using namespace std;

namespace
{
struct Course
{
int index, begin, end;
} C[100000];

int cmp(const void *p1, const void *p2)
{
struct Course *c1 = (struct Course*) p1;
struct Course *c2 = (struct Course*) p2;
if (c1->begin != c2->begin)
return c1->begin - c2->begin;
else
return c1->end - c2->end;
}
}

int main()
{
int N;
vector<vector<int> > V;
while (scanf("%d", &N) != EOF)
{
for (int i = 0; i < N; i++)
{
C[i].index = i + 1;
scanf("%d %d", &C[i].begin, &C[i].end);
}
qsort(C, N, sizeof(C[0]), cmp);

V.clear();
vector<int> v;
v.push_back(C[0].index);
V.push_back(v);
int end = C[0].end;
for (int i = 1; i < N; i++)
{
if (C[i].end <= end)
{
end = C[i].end;
V.back().push_back(C[i].index);
}
else if (C[i].begin >= end)
{
end = C[i].end;
vector<int> v;
v.push_back(C[i].index);
V.push_back(v);
}
else
{
V.back().push_back(C[i].index);
}
}

printf("%d\n", V.size());
for (size_t i = 0; i < V.size(); i++)
{
for (size_t j = 0; j < V[i].size(); j++)
printf(j ? " %d" : "%d", V[i][j]);
putchar('\n');
V[i].clear();
}
putchar('\n');
}
return 0;
}

#include<cstdio>
#include<cstdlib>

using namespace std;

namespace
{
struct Course
{
int index, begin, end;
} C[100000];

int cmp(const void *p1, const void *p2)
{
struct Course *c1 = (struct Course*) p1;
struct Course *c2 = (struct Course*) p2;
if (c1->begin != c2->begin)
return c1->begin - c2->begin;
else
return c1->end - c2->end;
}
}

int main()
{
int N, S[100000];
while (scanf("%d", &N) != EOF)
{
for (int i = 0; i < N; i++)
{
C[i].index = i + 1;
scanf("%d %d", &C[i].begin, &C[i].end);
}
qsort(C, N, sizeof(C[0]), cmp);

int group = 0;
int end = C[0].end;
for (int i = 1; i < N; i++)
{
if (C[i].end <= end)
end = C[i].end;
else if (C[i].begin >= end)
{
end = C[i].end;
S[group++] = i;
}
}
S[group++] = N;

printf("%d\n", group);
for (int i = 0; i < S[0]; i++)
printf(i == S[0] - 1 ? "%d\n" : "%d ", C[i].index);
for (int i = 1; i < group; i++)
for (int j = S[i - 1]; j < S[i]; j++)
printf(j == S[i] - 1 ? "%d\n" : "%d ", C[j].index);
puts("");
}
return 0;
}