http://poj.org/problem?id=2373
题意：
在长为L的草地上装喷水头，喷水头的喷洒半径为[a,b]
要求草地的每个整点被且只被一个喷水头覆盖
有N个特殊区间，要求只能被某一个喷水头完整地覆盖，而不能由多个喷水头分段覆盖
求喷水头的最小数目
喷水头只能建在整数点上
f[i] 表示区间[0,i]完全被覆盖的最少喷水头数目
f[i]=min(f[j]+1) j∈[i-2*a,i-2*b]
若i处于某个特殊区间内部，则f[i]=inf
单调队列优化
注意因为喷水头只能建在整数点上
当i为奇数时，区间[0,i]一共有偶数个点，喷水头不可能建在整数点上
所以只有i为偶数时才是有用的
这题真心不好写！！！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a,b;
int dp[1000001];
bool all[1000001];
struct node
{
int l,r;
}e[1001];
int q[1000001];
int h,t;
void read(int &x)
{
x=0; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); }
}
bool cmp(node p,node q)
{
if(p.l!=q.l) return p.l<q.l;
return p.r<q.r;
}
void up(int i)
{
while(h<t && i-q[h]>2*b) h++;
if(h<t && i-q[h]>=2*a)
{
dp[i]=dp[q[h]]+1;
while(h<t && dp[i]<dp[q[t-1]]) t--;
q[t++]=i;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("data.in","r",stdin);
freopen("xxy.out","w",stdout);
#endif
int n,l;
read(n); read(l);
read(a); read(b);
for(int i=1;i<=n;++i) read(e[i].l),read(e[i].r);
sort(e+1,e+n+1,cmp);
memset(dp,63,sizeof(dp));
int INF=dp[0];
dp[0]=0;
q[t++]=0;
int now=1;
int i=1;
while(i<=l)
{
if(now<=n && e[now].l<i)
{
if(e[now].r>i) i=e[now].r;
now++;
continue;
}
if(!(i&1)) up(i);
i++;
}
if(dp[l]==INF) printf("-1");
else printf("%d",dp[l]);
}
Dividing the Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5148 | Accepted: 1803 |
Description
Farmer John‘s cows have discovered that the clover growing along the ridge of the hill in his field is particularly good. To keep the clover watered, Farmer John is installing water sprinklers along the ridge of the hill.
To make installation easier, each sprinkler head must be installed along the ridge of the hill (which we can think of as a one-dimensional number line of length L (1 <= L <= 1,000,000); L is even).
Each sprinkler waters the ground along the ridge for some distance in both directions. Each spray radius is an integer in the range A..B (1 <= A <= B <= 1000). Farmer John needs to water the entire ridge in a manner that covers each location on the ridge by exactly one sprinkler head. Furthermore, FJ will not water past the end of the ridge in either direction.
Each of Farmer John‘s N (1 <= N <= 1000) cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval (S,E). Each of the cow‘s preferred ranges must be watered by a single sprinkler, which might or might not spray beyond the given range.
Find the minimum number of sprinklers required to water the entire ridge without overlap.
Input
* Line 1: Two space-separated integers: N and L
* Line 2: Two space-separated integers: A and B
* Lines 3..N+2: Each line contains two integers, S and E (0 <= S < E <= L) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge and so are in the range 0..L.
Output
* Line 1: The minimum number of sprinklers required. If it is not possible to design a sprinkler head configuration for Farmer John, output -1.
Sample Input
2 8 1 2 6 7 3 6
Sample Output
3
Hint
INPUT DETAILS:
Two cows along a ridge of length 8. Sprinkler heads are available in integer spray radii in the range 1..2 (i.e., 1 or 2). One cow likes the range 3-6, and the other likes the range 6-7.
OUTPUT DETAILS:
Three sprinklers are required: one at 1 with spray distance 1, and one at 4 with spray distance 2, and one at 7 with spray distance 1. The second sprinkler waters all the clover of the range like by the second cow (3-6). The last sprinkler waters all the clover of the range liked by the first cow (6-7). Here‘s a diagram:
|-----c2----|-c1| cows‘ preferred ranges
|---1---|-------2-------|---3---| sprinklers
+---+---+---+---+---+---+---+---+
0 1 2 3 4 5 6 7 8
The sprinklers are not considered to be overlapping at 2 and 6.
原文地址：https://www.cnblogs.com/TheRoadToTheGold/p/8386156.html