2017百度之星初赛(B)-1006-小小粉丝度度熊 hdu 6119
去重合并，以某个点为起点，向后二分到最后一个满足未签到天数<=m的点。
二分时的查询有很多方法，可以用线段树，dp，前缀数组等等。比赛时没有想到前缀数组，就用来dp来完成查询，查询一次大约O(logn)。前缀数组的话查询O(1)更快一些。
同学说尺取法也可以。
dp[i][j]代表第i个区间到第i+(1<<j)个区间的间隙和
#include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<iostream> #include<queue> #include<map> #include<cmath> #include<set> #include<stack> #define ll long long #define pb push_back #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)>(y)?(y):(x)) #define cls(name,x) memset(name,x,sizeof(name))//0或-1 #define fs first #define sc second #define mp make_pair #define L(x) (1<<x) #define next Next using namespace std; const int inf=1e9+10; const ll llinf=1e16+10; const int maxn=1e5+10; const int maxm=1e3+10; const int mod=1e9+7; int n,m; struct node { int l,r; }p[maxn]; bool cmp(const node &a,const node &b) { return a.l<b.l; } int dp[maxn][20]; void init() { for(int j=1;L(j)<=n;j++) for(int i=1;i+L(j)<=n;i++) dp[i][j]=dp[i][j-1]+dp[i+L(j-1)][j-1]; } int query(int L,int R) { if(L>=R) return 0; int t=0; while(1) { if(L+L(t+1)<=R) t++; else break; } return dp[L][t]+query(L+L(t),R); } int main() { //freopen("in.txt","r",stdin); while(~scanf("%d %d",&n,&m)) { for(int i=1;i<=n;i++) scanf("%d %d",&p[i].l,&p[i].r); sort(p+1,p+1+n,cmp); int k=1; for(int i=2;i<=n;i++) { if(p[i].l<=p[k].r+1&&p[i].r>p[k].r) p[k].r=p[i].r; else if(p[i].l>p[k].r+1) p[++k]=p[i]; } n=k; /*for(int i=1;i<=k;i++) printf("%d %d\n",p[i].l,p[i].r);*/ dp[n][0]=0; for(int i=1;i<=n-1;i++) dp[i][0]=p[i+1].l-p[i].r-1; init(); int ans=m; for(int i=1;i<=n-1;i++) { int l=i+1,r=n; int temp=i; while(l<=r) { int mid=(l+r)/2; if(query(i,mid)<=m) { temp=max(temp,mid); l=mid+1; } else r=mid-1; } ans=max(ans,p[temp].r-p[i].l+1+m-query(i,temp)); } ans=max(ans,p[n].r-p[n].l+1+m); printf("%d\n",ans); } return 0; }
sum[i]代表第i个区间之前的所有间隙和
#include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<iostream> #include<queue> #include<map> #include<cmath> #include<set> #include<stack> #define ll long long #define pb push_back #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)>(y)?(y):(x)) #define cls(name,x) memset(name,x,sizeof(name))//0或-1 #define fs first #define sc second #define mp make_pair #define L(x) (1<<x) #define next Next using namespace std; const int inf=1e9+10; const ll llinf=1e16+10; const int maxn=1e5+10; const int maxm=1e3+10; const int mod=1e9+7; int n,m; struct node { int l,r; }p[maxn]; bool cmp(const node &a,const node &b) { return a.l<b.l; } int sum[maxn]; int query(int L,int R) { return sum[R]-sum[L]; } int main() { //freopen("in.txt","r",stdin); while(~scanf("%d %d",&n,&m)) { for(int i=1;i<=n;i++) scanf("%d %d",&p[i].l,&p[i].r); sort(p+1,p+1+n,cmp); int k=1; for(int i=2;i<=n;i++) { if(p[i].l<=p[k].r+1 && p[i].r>p[k].r) p[k].r=p[i].r; else if(p[i].l>p[k].r+1) p[++k]=p[i]; } n=k; sum[1]=0; for(int i=2;i<=n;i++) sum[i]=sum[i-1]+p[i].l-p[i-1].r-1; int ans=m; for(int i=1;i<=n-1;i++) { int l=i+1,r=n; int temp=i; while(l<=r) { int mid=(l+r)/2; if(query(i,mid)<=m) { temp=max(temp,mid); l=mid+1; } else r=mid-1; } ans=max(ans,p[temp].r-p[i].l+1+m-query(i,temp)); } ans=max(ans,p[n].r-p[n].l+1+m); printf("%d\n",ans); } return 0; }
尺取法
#include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<iostream> #include<queue> #include<map> #include<cmath> #include<set> #include<stack> #define ll long long #define pb push_back #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)>(y)?(y):(x)) #define cls(name,x) memset(name,x,sizeof(name))//0或-1 #define fs first #define sc second #define mp make_pair #define L(x) (1<<x) #define next Next using namespace std; const int inf=1e9+10; const ll llinf=1e16+10; const int maxn=1e5+10; const int maxm=1e3+10; const int mod=1e9+7; int n,m; struct node { int l,r; }p[maxn]; bool cmp(const node &a,const node &b) { return a.l<b.l; } int main() { //freopen("in.txt","r",stdin); while(~scanf("%d %d",&n,&m)) { for(int i=1;i<=n;i++) scanf("%d %d",&p[i].l,&p[i].r); sort(p+1,p+1+n,cmp); int k=1; for(int i=2;i<=n;i++) { if(p[i].l<=p[k].r+1 && p[i].r>p[k].r) p[k].r=p[i].r; else if(p[i].l>p[k].r+1) p[++k]=p[i]; } n=k; int l=1,r=1,cost=0; int ans=m; while(l<=n-1) { while(p[r+1].l-p[r].r-1 <= m-cost && r+1<=n) { cost+=p[r+1].l-p[r].r-1; r++; } ans=max(ans,p[r].r-p[l].l+1 + m-cost); if(l!=r) cost-=p[l+1].l-p[l].r-1; l++; r=max(r,l); } ans=max(ans,p[n].r-p[n].l+1 + m); printf("%d\n",ans); } return 0; }
