差分约束系统 ZQUOJ 23147&&POJ 1201 Intervals

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：构造一个集合，这个集合内的数字满足所给的n个条件，每个条件都是指在[a,b]内至少有c个数在集合内。问这个集合最少包含多少个点。即求至少有多少个元素在区间[a,b]内。
分析：用ti表示在[0,i-1]的范围内，要选多少个数，对于题目中所说的每个条件[a,b]内至少有c个数在集合可以表示为t(b+1)-t(a)>=c,差分约束系统的雏形已经形成。但是，题目还有一个隐藏的条件，ti表示的是在[0,i-1]的范围内，要选多少个数，数列ti是递增的，但是增量最大是1，也就是说0<=t(i)-t(i-1)<=1，这个式子等价于t(i)-t(i-1)<=1和t(i-1)-t(i)<=0。用spfa算法得到一个最长路，第一个到最后一个节点的最长路即是需要求的值。
AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 1000000
using namespace std;
typedef struct
{
    int v,w,next;
}Node;
Node e[250000];
int first[50010],vis[50010],dist[50010];
int n,g,minx,maxx;
void add(int u,int v,int w)
{
    e[g].v=v;
    e[g].w=w;
    e[g].next=first[u];
    first[u]=g++;
}
void SPFA()
{
    int i,j,t;
    queue<int> q;
    for(i=minx;i<=maxx;i++)
    {
        vis[i]=0;
        dist[i]=-INF;
    }
    while(!q.empty())
        q.pop();
    dist[minx]=0;
    vis[minx]=1;
    q.push(minx);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        vis[t]=0;
        for(i=first[t];i!=-1;i=e[i].next)
        {
            j=e[i].v;
            if(dist[j]<dist[t]+e[i].w)
            {
                dist[j]=dist[t]+e[i].w;
                if(!vis[j])
                {
                    vis[j]=1;
                    q.push(j);
                }
            }
        }
    }
}
int main()
{
    int ai,bi,ci,i;
    scanf("%d",&n);
    memset(first,-1,sizeof(first));
    g=0; minx=INF; maxx=0;
    for(i=1;i<=n;i++)
    {
        scanf("%d%d%d",&ai,&bi,&ci);
        if(ai<minx)
            minx=ai;
        if(bi+1>maxx)
            maxx=bi+1;
        add(ai,bi+1,ci);
    }
    for(i=minx;i<=maxx;i++)
    {
        add(i,i-1,-1);
        add(i-1,i,0);
    }
    SPFA();
    printf("%d\n",dist[maxx]);
    return 0;
}