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POJ1201-Intervals(差分约束)
Intervals
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 20786 | Accepted: 7866 |
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
5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1
Sample Output
6
Source
Southwestern Europe 2002
一道典型的差分约束:
题目意思:
给你m个区间,每个区间至少要取c个数
问最少取多少个数
解法:
用X(i) 表示前i个数中取了多少个数
对于每个区间的约束建立下列不等不等式:
X(a) - X(b+1) <= -c
除此之外还有补充另外的边
X(i+1)-X(i) <= 1
要求的是
X(sink) - X(src) >= d (d即为所求)
即
X(src)-X(sink) <= -d
求sink到src的最短路径
SPFA搞定
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> using namespace std; const int maxn = 50000+10000; const int INF = 1e9; int n; int src,sink; bool inQue[maxn]; int dist[maxn]; queue<int>que; struct edge{ int to,next,w; edge(int to,int next,int w):to(to),next(next),w(w){} }; int head[maxn]; vector<edge> e; void addedge(int from,int to,int w){ e.push_back(edge(to,head[from],w)); head[from] = e.size()-1; } void spfa(){ for(int i = src; i <= sink; i++){ inQue[i] = 0; dist[i] = INF; } inQue[sink] = 1; que.push(sink); dist[sink] = 0; while(!que.empty()){ int u = que.front(); que.pop(); inQue[u] = 0; for(int i = head[u]; i != -1; i = e[i].next){ if(dist[e[i].to] > dist[u]+e[i].w){ dist[e[i].to] = dist[u]+e[i].w; if(!inQue[e[i].to]){ inQue[e[i].to] = 1; que.push(e[i].to); } } } } } int main(){ int m; freopen("in","r",stdin); while(~scanf("%d",&m)){ e.clear(); src = http://www.mamicode.com/maxn,sink = -1;>
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