dilworth解决的题目poj3636,poj1631

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//最小的链划分数问题。根据dilworth定理//为最大的反链的长度//187MS 1860K #include <iostream>#include <algorithm>using namespace std;static const int INF = ~0U>>1; static struct node{int w;int h;bool operator<(node&rs){return w<rs.w||(w==rs.w&&h>rs.h);}}Nodes[20005];static int n;static int binsearch(int B[],int num,int l,int r){int mid;while(l<r){mid = (l+r)/2;if (num<=B[mid])//下降子序列 l = mid+1;else r = mid;}return l;}//用动规求解static void DP(){int B[20010];memset(B,0,sizeof(B));//求最长下降子序列int l=0,r;int len = 0;r = len;for (int i=0;i<n;++i){l = 0;r=len;int index  = binsearch(B,Nodes[i].h,l,r);B[index] = Nodes[i].h;if (len==index) len++;}printf("%d\n",len);}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);for (int i=0;i<n;++i){scanf("%d %d",&Nodes[i].w,&Nodes[i].h);}sort(Nodes,Nodes+n);DP();}return 0;}

//输出最长没有交叉//109MS 2072K //就是求最长上升子序列#include <iostream>#include <algorithm>using namespace std;static const int INF = ~0U>>1;  static struct _node{int l,r;bool operator<(_node&b){return l<b.l||(l==b.l&&r<b.r);}}Nodes[40005];static int n;static int binsearch(int B[],int num,int l,int r){int mid;while(l<r){mid = (l+r)/2;if (num<=B[mid])//上升子序列r = mid;elsel = mid+1;}return l;}//用动规求解static void DP(){int B[40005];memset(B,0,sizeof(B));//求最长下降子序列int l=0,r;int len = 0;r = len;for (int i=0;i<n;++i){B[i] = INF;l = 0;r=len;int index  = binsearch(B,Nodes[i].r,l,r);B[index] = Nodes[i].r;if (len==index)len++;}printf("%d\n",len);}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);for (int i=0;i<n;++i){scanf("%d",&Nodes[i].r);Nodes[i].l = i+1;}sort(Nodes,Nodes+n);DP();}return 0;}

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