Description
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
Output
Sample Input
5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1
Sample Output
6
Source
/**
題意:給出n個閉區間[ai,bi],每個區間對應一個ci,
表示集合Z在區間[ai,bi]內ci個相同元素,
問集合Z至少有幾個元素。
分析 :差分約束入門。
如果我用Si表示區間[0,i]區間內至少有多少個元素的話,
那麼Sbi - Sai >= ci,這樣我們就構造出來了一系列邊,權值爲ci,
但是這遠遠不夠,因爲有很多點依然沒有相連接起來,
不難寫出0<=Si - Si-1<=1的限制條件。
我們將上面的限制條件寫爲同意的形式:
Sbi - Sai >= ci
Si - Si-1 >= 0
Si-1 - Si >= -1
對於 b - a > c , addedge(a,b,c);
*/
#include <stdio.h>
#include <algorithm>
#include<cstring>
#include <iostream>
#include<queue>
using namespace std;
const int N = 50010;
const int INF = 1<<30;
int mn = INF,mx = -1;
int d[N],first[N],v[N<<2],w[N<<2],next[N<<2];
char inq[N];
int p = 0;
void Bellman_Ford()
{
for(int i=mn; i<=mx; i++)d[i]=-INF;
d[mn] = 0;
queue<int>q;
q.push(mn);
while(!q.empty())
{
int u = q.front();
q.pop();
inq[u]=0;
for(int i=first[u]; i!=-1; i=next[i])
{
if(d[v[i]] < d[u]+w[i])
{
d[v[i]] = d[u]+w[i];
if(!inq[v[i]])
{
inq[v[i]]=1;
q.push(v[i]);
}
}
}
}
}
void addedge(int a, int b, int k)
{
v[++p] = b;
next[p] = first[a];
first[a] = p;
w[p] = k;
}
int main()
{
//freopen("date.in","r",stdin);
int i,j,k,m,n,a,b;
memset(first,-1,sizeof(first));
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%d %d %d",&a,&b,&k);
b++;
mn = min(mn,a), mx = max(mx,b);
addedge(a,b,k);
}
for(i=mn; i<mx; i++)
addedge(i,i+1,0), addedge(i+1,i,-1);
Bellman_Ford();
printf("%d\n",d[mx]);
}