POJ1201 Intervals(差分约束)

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28416		Accepted: 10966

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

Source

Southwestern Europe 2002

 

 

题意：每次给出一段区间$[a_i,b_i]$以及一个数$c_i$，使得在这中间至少有$c_i$个数，求一个最小的集合$Z$，使得集合$Z$满足上述所有要求，问集合$Z$的大小

 

 

 

思路：

设$S[i]$表示$0-i$这一段区间的前缀和

那么题目的关系就变成了$S[b_i]-S[a_i]>=c_i$

这是一个很典型的差分约束类问题

题目中要求集合最小，因此转换为最长路，将所有的式子写成$B-A>=C$的形式

同时题目中还有一个条件$0<=S[i]-S[i-1]<=1$

因为数据为整数

于是又得到两个方程

$Sleft[ i ight] -Sleft[ i-1 ight] geq 0$
$Sleft[ i-1 ight] -Sleft[ i ight] geq -1$

但是有个细节：$S[i-1]$不能表示，因此我们需要将所有下标$+1$，此时$S[i]$表示$0 to (i-1)$的前缀和

同时，这个图一定是联通的，因此不用新建超级源点

 

#include<cstdio>
#include<queue>
#include<cstring>
#define INF 1e8+10
using namespace std;
const int MAXN=1e6+10;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++)
char buf[MAXN],*p1=buf,*p2=buf;
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
struct node
{
    int u,v,w,nxt;
}edge[MAXN];
int head[MAXN],num=1;
int maxx=-INF,minn=INF;
int dis[MAXN],vis[MAXN];
inline void AddEdge(int x,int y,int z)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].w=z;
    edge[num].nxt=head[x];
    head[x]=num++;
}
int SPFA()
{
    queue<int>q;
    memset(dis,-0xf,sizeof(dis));
    dis[minn]=0;q.push(minn); 
    while(q.size()!=0)
    {
        int p=q.front();q.pop();
        vis[p]=0;
        for(int i=head[p];i!=-1;i=edge[i].nxt)
        {
            if(dis[edge[i].v]<dis[p]+edge[i].w)
            {
                dis[edge[i].v]=dis[p]+edge[i].w;
                if(vis[edge[i].v]==0)
                    vis[edge[i].v]=1,q.push(edge[i].v);
            }
        }
    }
    printf("%d",dis[maxx]);
}
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    memset(head,-1,sizeof(head));
    int N=read();
    for(int i=1;i<=N;i++)
    {
        int x=read(),y=read(),z=read();
        AddEdge(x,y+1,z); 
        maxx=max(y+1,maxx);
        minn=min(x,minn);
    }
    for(int i=minn;i<=maxx-1;i++)
    {
        AddEdge(i+1,i,-1);
        AddEdge(i,i+1,0);
    }
    SPFA();
    return 0;
}