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[poj 1201]Intervals 差分约束

来源:程序员人生   发布时间:2016-06-07 17:19:14 阅读次数:2196次

题目:http://poj.org/problem?id=1201

Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 24502 Accepted: 9317

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.

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
5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

Sample Output
6

题意:要求满足bi-ai选数>=ci 中最小的集合;

思路
如题:s[n]代表前n当选最小的集合,可得:

s[bi]-s[ai-1]>=ci;(题) s[I+1]-s[I]>=0;(本身) s[I+1]-s[I]<=1;-->s[l]-s[I+1]>=-1(本身)

用最小值建边spfa求maxn-minn的最长路便可;

代码

#include<iostream> #include<stdio.h> #include<string.h> #include<queue> using namespace std; int head[60005],val[160005]; int tot,to[160005],nex[160005]; int n; int aa,bb,cc; queue<int> q; int d[60005]; int vis[60005]; void spfa(int x) { memset(d,-1,sizeof(d)); d[x]=0; vis[x]=1; q.push(x); while(!q.empty()) { int now=q.front(); q.pop(); vis[now]=0; for(int i=head[now];i!=-1;i=nex[i]) { int v=to[i]; if(d[v]==-1||d[v]<d[now]+val[i]) { d[v]=d[now]+val[i]; if(!vis[v]) { vis[v]=1; q.push(v); } } } } } void add(int x,int y,int v) { int tmp=head[x]; head[x]=++tot; nex[tot]=tmp; to[tot]=y; val[tot]=v; } int main() { scanf("%d",&n); int minn=10000000; int maxn=-10000000; memset(head,-1,sizeof(head)); for(int i=1;i<=n;i++) { scanf("%d%d%d",&aa,&bb,&cc); aa--; add(bb,aa,cc); minn=min(aa,minn); maxn=max(maxn,bb); } for(int i=minn;i<maxn;i++) { add(i+1,i,0); add(i,i+1,-1); } spfa(maxn); printf("%d\n",d[minn]); }
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