Intervals
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 20779 | Accepted: 7863 |
Description
You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.
Write a program that:
reads the number of intervals, their end points and integers c1, ..., cn from the standard input,
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n,
writes the answer to the standard output.
Write a program that:
reads the number of intervals, their end points and integers c1, ..., cn from the standard input,
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n,
writes the answer to the standard output.
Input
The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <=
ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.
ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.
Output
The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.
Sample Input
5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
Sample Output
6
Source
题目大意：
n行。每行a,b,c，表示在区间a,b内要找c个数，问你总共至少要找多少个数？
解题思路：
差分约束系统。
在本题中。假设[a,b]中要找c个元素。那么：s[b]-s[a-1]>=c。我们能够推得：s[a-1] - s[b] <= -c
同一时候。因为每个值上最多仅仅能含有一个元素。那么:s[i] - s[i-1]<=1 ，又因为s[i] - s[i-1]>=0 推得：s[i-1] - s[i] <=0
这样：我们有了三个约束不等式：
s[a-1] - s[b] <= -c
s[i] - s[i-1]<=1
s[i-1] - s[i] <=0于是：假设起点为from，终点为to。我们仅仅要求出：s[to] -s[from-1] >= M就能够了，
因此求出 s[from-1]-s[to]<=-M，即求to 到 from-1 的最短路径，
注意：因为i<0，所以i-1可能小于0，因此所有向右平移1位。
解题代码：
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std; const int maxn=50000;
const int inf=0x3f3f3f3f; struct edge{
int u,v,w,next;
}e[maxn*10]; int head[maxn*2+10],dist[maxn*2+10],cnt;
int n,from,to; void initial(){
cnt=0;
from=inf,to=0;
for(int i=0;i<=maxn;i++) head[i]=-1;
} void adde(int u,int v,int w){
u++;v++;
e[cnt].u=u,e[cnt].v=v,e[cnt].w=w,e[cnt].next=head[u],head[u]=cnt++;
} void input(){
int u,v,w;
//[v]-[u-1]>=w [u-1]-[v]<=-w
for(int i=0;i<n;i++){
scanf("%d%d%d",&u,&v,&w);
adde(v,u-1,-w);
if(u<from) from=u;
if(v>to) to=v;
}
//0<=[i]-[i-1]<=1
for(int i=from;i<=to;i++){
adde(i-1,i,1);
adde(i,i-1,0);
}
} bool spfa(int from){
int s=from,num[maxn];
bool visited[maxn];
for(int i=0;i<=maxn;i++){
num[i]=0;
dist[i]=inf;
visited[i]=false;
}
queue <int> q;
q.push(s);
visited[s]=true;
dist[s]=0;
while(!q.empty()){
s=q.front();
q.pop();
for(int i=head[s];i!=-1;i=e[i].next){
int d=e[i].v;
if(dist[d]>dist[s]+e[i].w){
dist[d]=dist[s]+e[i].w;
if(!visited[d]){
visited[d]=true;
q.push(d);
num[d]++;
if(num[d]>n) return false;
}
}
}
visited[s]=false;
}
return true;
} void solve(){
//get [to]-[from-1]>=M; [from-1]-[to]<=-M
spfa(to+1);
cout<<-dist[from]<<endl;
} int main(){
while(scanf("%d",&n)!=EOF){
initial();
input();
solve();
}
return 0;
}