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这个题很显然，可以转换成这个问题：为什么把上限最小的区间弹出优先考虑呢？因为在有解的情况下，对于第i个数而言，堆中所有区间都可以包含这个数，且区间下限都会被第j(j>i)个数包含，所以现在只需要考虑区间上限这一个因素，考虑贪心的原则，很显然要优先把上限最小的区间先处理掉。
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
struct Island
{
ll l,r;
int no;
bool operator <(Island one)const
{
return r>one.r;
}
}island[int(2e5)+10];
bool cmp(Island one,Island two)
{
return one.l<two.l;
}
struct Bridge
{
ll len;
int no;
}bridge[int(2e5)+10];
bool cmp1(Bridge one,Bridge two)
{
return one.len<two.len;
}
priority_queue<Island>qq;
int ans[int(2e5)+10];
int main()
{
int n,m;
cin>>n>>m;
ll pl,pr;
for(int i=0;i<n;i++)
{
ll l,r;
cin>>l>>r;
if(i>0)
{
island[i-1].l=l-pr;
island[i-1].r=r-pl;
island[i-1].no=i-1;
}
pl=l;
pr=r;
}
sort(island,island+n-1,cmp);
for(int i=0;i<m;i++)
{
cin>>bridge[i].len;
bridge[i].no=i;
}
sort(bridge,bridge+m,cmp1);
int sum=0;
for(int i=0,j=0;i<m;i++)
{
while(j<n-1&&island[j].l<=bridge[i].len&&bridge[i].len<=island[j].r)
qq.push(island[j++]);
if(qq.empty())
continue;
Island t=qq.top();
qq.pop();
if(bridge[i].len>t.r)
{
puts("No");
return 0;
}
ans[t.no]=bridge[i].no;
sum++;
}
if(sum<n-1)
{
puts("No");
return 0;
}
puts("Yes");
for(int i=0;i<n-1;i++)
{
if(i)
cout<<" ";
cout<<ans[i]+1;
}
}Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let‘s represent them as non-intersecting segments on a straight line: island i has coordinates [li,?ri], besides, ri?<?li?+?1 for 1?≤?i?≤?n?-?1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i?+?1)-th islads, if there are such coordinates of x and y, that li?≤?x?≤?ri, li?+?1?≤?y?≤?ri?+?1 and y?-?x?=?a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
The first line contains integers n (2?≤?n?≤?2·105) and m (1?≤?m?≤?2·105) — the number of islands and bridges.
Next n lines each contain two integers li and ri (1?≤?li?≤?ri?≤?1018) — the coordinates of the island endpoints.
The last line contains m integer numbers a1,?a2,?...,?am (1?≤?ai?≤?1018) — the lengths of the bridges that Andrewid got.
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n?-?1 numbers b1,?b2,?...,?bn?-?1, which mean that between islands i and i?+?1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
4 4 1 4 7 8 9 10 12 14 4 5 3 8
Yes 2 3 1
2 2 11 14 17 18 2 9
No
2 1 1 1 1000000000000000000 1000000000000000000 999999999999999999
Yes 1
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn‘t exist.
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codeforces 556 D Case of Fugitive
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原文地址：http://blog.csdn.net/stl112514/article/details/46868749
