POJ1201 Intervals （差分约束）

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：给定X轴上一些区间[ai，bi]，其间至少有ci个点，求X轴上至少多少个点。

思路：记得高中是用bellman-ford优化到了NKOJ的第一名。。。。但是差不多搞忘差分约束了。所以现在再来整理总结一下。

此处求最小，显然需要构造T>=S+dist，然后用最长路求。 由于是点段，处理区间最好半开半闭。

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1500000;
const int inf=0x7fffffff;
int cnt,n,Max,Min;
int Laxt[maxn],Next[maxn],To[maxn],Len[maxn];
int times[maxn],st[maxn],vis[maxn],inq[maxn];
void update()
{
    cnt=0;Min=inf;Max=-inf;
    memset(Laxt,0,sizeof(Laxt));
    memset(vis,0,sizeof(vis));
    memset(inq,0,sizeof(inq));
}
void add(int u,int v,int d)
{
    Next[++cnt]=Laxt[u];
    Laxt[u]=cnt;
    To[cnt]=v;
    Len[cnt]=d;
}
bool spfa()
{
    for(int i=Min-1;i<=Max+1;i++)  st[i]=-inf;
    queue<int>q;
    q.push(Min); st[Min]=0; inq[Min]=1;
    while(!q.empty()){
        int u=q.front(); q.pop(); inq[u]=0;
        for(int i=Laxt[u];i;i=Next[i]){
            int v=To[i];
            if(st[v]<st[u]+Len[i]){
                st[v]=st[u]+Len[i];
                if(!inq[v]){
                   inq[v]=1; vis[v]++; q.push(v);
                   //if(vis[v]>Max-Min) return false;
                }
            }
        }
    } return true;
}
int main()
{
    int a,b,c;
    while(~scanf("%d",&n)){
        update();
        for(int i=1;i<=n;i++) {
            scanf("%d%d%d",&a,&b,&c);
            a++;b++;
            if(b>Max) Max=b;
            if(a-1<Min) Min=a-1;
            add(a-1,b,c);
        }
        for(int i=Min;i<Max;i++) add(i+1,i,-1),add(i,i+1,0);
        spfa();
        printf("%d\n",st[Max]);
    } return 0;
}

原文地址：https://www.cnblogs.com/hua-dong/p/8438131.html