题目：

Covering the Corral

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Description

The cows are so modest they want Farmer John to install covers around the circular corral where they occasionally gather. The corral has circumference C (1 ≤ C ≤ 1,000,000,000), and FJ can choose from a set of M (1 ≤ M ≤100,000) covers that have fixed starting points and sizes. At least one set of covers can surround the entire corral.

Cover i can be installed at integer location x_i (distance from starting point around corral) (0 ≤ x_i < C) and has integer length l_i (1 ≤ l_i ≤ C).

FJ wants to minimize the number of segments he must install. What is the minimum number of segments required to cover the entire circumference of the corral?

Consider a corral of circumference 5, shown below as a pair of connected line segments where both '0's are the same point on the corral (as are both 1's, 2's, and 3's).

Three potential covering segments are available for installation:

Start Length
i x_i l_i
1 0 1
2 1 2
3 3 3

Installing segments 2 and 3 cover an extent of (at least) five units around the circumference. FJ has no trouble with the overlap, so don't worry about that.

Input

* Line 1: Two space-separated integers: C and M

* Lines 2..M+1: Line i+1 contains two space-separated integers: x_i and l_i

Output

* Line 1: A single integer that is the minimum number of segments required to cover all segments of the circumference of the corral

Sample Input

5 3
0 1
1 2
3 3

Sample Output

题意：

给出一个10亿的环，再给出10W个固定区间，问最少选取多少个区间能够覆盖完整个环？

思路：

首先按区间起点排序，然后贪心找出选了每一个区间之后，下一个区间应该选谁

注意到当遇到某个区间的y不递增的时候，这个区间显然会被别的区间所完全覆盖，那么这个区间是一个无效区间，所以有效区间的y也是递增的，那么我们就可以使用单调队列了，

然后是倍增法，dp[i][j]表示选取第i个区间的时候接下来第2^j个区间要取谁

那么有状态转移：dp[i][j]=dp[dp[i][j-1]][j-1]

倍增大法好，但是貌似被我写搓了，看了下别人的代码，有效率是我1/4的_(:з」∠)_

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXN 100010
#define inf 0x3f3f3f3f
struct node
{long long x,z;int next;
}k[MAXN*2];
struct node2
{long long x,z;int i;
}dl[MAXN*2];
int n,m,up;
int mi[25];
int dp[MAXN*3][17];
int cmp(const void *a,const void *b)
{return (*(struct node *)a).x>(*(struct node *)b).x?1:-1;
}
void find_next()
{int i,j,l,s=0,t=-1,lastz=-1,sign=1;for(i=0;i<n;i++){if(k[i].z<=lastz)///特判无用区间{k[i].next=-1;continue;}for(l=s;l<=t;l++)if(k[i].x<dl[l].x)break;s=l;for(j=sign;k[j].x<=k[i].z+1&&j<n*2;j++){for(l=t;l>=s;l--)if(k[j].z<=dl[l].z)break;t=l+1;dl[t].z=k[j].z;dl[t].x=k[j].x;dl[t].i=j;}sign=j;if(t<s)k[i].next=-1;elsek[i].next=dl[s].i;lastz=k[i].z;}
}int pd(int x,long long sign)
{if(x==-1)return 1;if(k[x].x<sign&&k[x].z+1>=sign)return 0;if(k[x].x>=sign)return 1;elsereturn -1;
}
int find_ans(int t)
{///dp[i][j]代表第i个答案的下j^2个答案int i,j,res=1,tmp,top=up;long long sign=k[t+n].x;if(k[t].z+1>=sign)return 1;if(dp[t][0]==-1)return inf;while(1){for(j=0;j<=top;j++){tmp=pd(dp[t][j],sign);if(tmp==0)///当前答案恰好return res+mi[j];if(tmp==1)///当前答案太大break;}if(j==0||j==1)return inf;else{t=dp[t][j-1];res+=mi[j-1];top=j-2;}}
}
int main()
{int i,j,t,ans,tmp;mi[0]=1;for(i=1;i<=20;i++)mi[i]=mi[i-1]*2;while(scanf("%d%d",&m,&n)!=EOF){tmp=n;up=-1;do{tmp/=2;up++;}while(tmp);for(i=0;i<n;i++)scanf("%lld%lld",&k[i].x,&k[i].z);qsort(k,n,sizeof(k[0]),cmp);for(i=0;i<n;i++){k[i].z=k[i].x+k[i].z-1;k[i+n]=k[i];k[i+n].x+=m;k[i+n].z+=m;}find_next();memset(dp,-1,sizeof(dp));for(i=0;i<n;i++){dp[i][0]=k[i].next;if(dp[i][0]!=-1)dp[i+n][0]=k[i].next+n;}for(j=1;j<=up;j++)for(i=0;i<n;i++)if(dp[i][j-1]==-1)dp[i][j]=-1;elsedp[i][j]=dp[dp[i][j-1]][j-1];ans=inf;for(i=0;i<n;i++){if(dp[i][0]==-1)continue;t=find_ans(i);if(t<ans)ans=t;}printf("%d\n",ans);}return 0;
}

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DELL Software contest 2014 H题 1007 Covering the Corral

Covering the Corral

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