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

Sample Output

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

$S\left[ i\right] -S\left[ i-1\right] \geq 0$
$S\left[ i-1\right] -S\left[ i\right] \geq -1$

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+;

#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=,f=;

    while(c<''||c>''){if(c=='-')f=-;c=getchar();}

    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}

    return x*f;

}

struct node

{

    int u,v,w,nxt;

}edge[MAXN];

int head[MAXN],num=;

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]=;q.push(minn);

    while(q.size()!=)

    {

        int p=q.front();q.pop();

        vis[p]=;

        for(int i=head[p];i!=-;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]==)

                    vis[edge[i].v]=,q.push(edge[i].v);

            }

        }

    }

    printf("%d",dis[maxx]);

}

int main()

{

    #ifdef WIN32

    freopen("a.in","r",stdin);

    #else

    #endif

    memset(head,-,sizeof(head));

    int N=read();

    for(int i=;i<=N;i++)

    {

        int x=read(),y=read(),z=read();

        AddEdge(x,y+,z);

        maxx=max(y+,maxx);

        minn=min(x,minn);

    }

    for(int i=minn;i<=maxx-;i++)

    {

        AddEdge(i+,i,-);

        AddEdge(i,i+,);

    }

    SPFA();

    return ;

}

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