POJ 1201 Intervals（图论-差分约束）

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20779		Accepted: 7863

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

题目大意：

n行。每行a,b,c，表示在区间a,b内要找c个数，问你总共至少要找多少个数？

解题思路：

差分约束系统。
在本题中。假设[a,b]中要找c个元素。那么：s[b]-s[a-1]>=c。我们能够推得：s[a-1] - s[b] <= -c

同一时候。因为每个值上最多仅仅能含有一个元素。那么:s[i] - s[i-1]<=1 ，又因为s[i] - s[i-1]>=0 推得：s[i-1] - s[i] <=0

这样：我们有了三个约束不等式：
s[a-1] - s[b] <= -c
s[i] - s[i-1]<=1
s[i-1] - s[i] <=0

于是：假设起点为from，终点为to。我们仅仅要求出：s[to] -s[from-1] >= M就能够了，

因此求出 s[from-1]-s[to]<=-M，即求to 到 from-1 的最短路径，

注意：因为i<0，所以i-1可能小于0，因此所有向右平移1位。

解题代码：

#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;

const int maxn=50000;
const int inf=0x3f3f3f3f;

struct edge{
    int u,v,w,next;
}e[maxn*10];

int head[maxn*2+10],dist[maxn*2+10],cnt;
int n,from,to;

void initial(){
    cnt=0;
    from=inf,to=0;
    for(int i=0;i<=maxn;i++) head[i]=-1;
}

void adde(int u,int v,int w){
    u++;v++;
    e[cnt].u=u,e[cnt].v=v,e[cnt].w=w,e[cnt].next=head[u],head[u]=cnt++;
}

void input(){
    int u,v,w;
    //[v]-[u-1]>=w [u-1]-[v]<=-w
    for(int i=0;i<n;i++){
        scanf("%d%d%d",&u,&v,&w);
        adde(v,u-1,-w);
        if(u<from) from=u;
        if(v>to) to=v;
    }
    //0<=[i]-[i-1]<=1
    for(int i=from;i<=to;i++){
        adde(i-1,i,1);
        adde(i,i-1,0);
    }
}

bool spfa(int from){
    int s=from,num[maxn];
    bool visited[maxn];
    for(int i=0;i<=maxn;i++){
        num[i]=0;
        dist[i]=inf;
        visited[i]=false;
    }
    queue <int> q;
    q.push(s);
    visited[s]=true;
    dist[s]=0;
    while(!q.empty()){
        s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(dist[d]>dist[s]+e[i].w){
                dist[d]=dist[s]+e[i].w;
                if(!visited[d]){
                    visited[d]=true;
                    q.push(d);
                    num[d]++;
                    if(num[d]>n) return false;
                }
            }
        }
        visited[s]=false;
    }
    return true;
}

void solve(){
    //get [to]-[from-1]>=M; [from-1]-[to]<=-M
    spfa(to+1);
    cout<<-dist[from]<<endl;
}

int main(){
    while(scanf("%d",&n)!=EOF){
        initial();
        input();
        solve();
    }
    return 0;
}