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poj 1201 Intervals
Intervals
http://poj.org/problem?id=1201
| Time Limit: 2000MS | Memory Limit: 65536K | |
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.
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
53 7 38 10 36 8 11 3 110 11 1
Sample Output
6
Source
Southwestern Europe 2002
题目大意：给出n个区间[a,b]，每个区间指定要选出c个数，求满足所有的区间最少要选出多少个数
poj 1716的弱化版
在此不再详细解读
可参考 http://www.cnblogs.com/TheRoadToTheGold/p/6529744.html
#include<cstdio>#include<queue>#include<cstring>#include<algorithm>#define N 50001using namespace std;int n,minn=50002,maxn;queue<int>q;struct node{ int to,next,w;}e[N*3];int dis[N],front[N],tot;bool v[N];void add(int u,int v,int w){ e[++tot].to=v;e[tot].next=front[u];front[u]=tot;e[tot].w=w;}int main(){ scanf("%d",&n); int x,y,z; for(int i=1;i<=n;i++) { scanf("%d%d%d",&x,&y,&z);y++; add(x,y,z); minn=min(x,minn);maxn=max(maxn,y); } for(int i=minn;i<maxn;i++) { add(i,i+1,0);add(i+1,i,-1); } memset(dis,-1,sizeof(dis)); q.push(minn);v[minn]=true;dis[minn]=0; while(!q.empty()) { int now=q.front();q.pop();v[now]=false; for(int i=front[now];i;i=e[i].next) { int to=e[i].to; if(dis[to]<dis[now]+e[i].w) { dis[to]=dis[now]+e[i].w; if(!v[to]) { v[to]=true; q.push(to); } } } } printf("%d",dis[maxn]);}
poj 1201 Intervals
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