poj1201——差分约束，spfa

poj1201——差分约束，spfa

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22553		Accepted: 8530

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：给定N个区间[ai,bi],对应ci，求一个集合，对每个区间都有ci个数在集合中，求集合的数的最小个数
思路：差分约束，s(bi)-s(ai-1)>=ci,s(i+1)-s(i)>=0,s(i)-s(i+1)>=-1。以区间最左端的Min为源点，求最长路

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<queue>



using namespace std;



const int maxn=50010;

const int INF=(1<<29);



int N;

int a,b,c;

struct Edge

{

    int v,w;

    Edge *next;

};Edge e[maxn*10];

bool vis[maxn];

int cnt[maxn];

int dist[maxn];

int Min,Max;///Min作为源点



void add_edge(int u,int v,int w)

{

    Edge *pre=&e[u];

    Edge *p=(Edge*)malloc(sizeof(Edge));

    p->v=v;p->w=w;

    p->next=pre->next;

    pre->next=p;

}



bool relax(int u,int v,int w)

{

    if(dist[u]+w>dist[v]){

        dist[v]=dist[u]+w;

        return true;

    }

    return false;

}



bool spfa()

{

    for(int i=Min;i<=Max+1;i++) dist[i]=-INF;

    dist[Min]=0;

    memset(vis,0,sizeof(vis));

    memset(cnt,0,sizeof(cnt));

    queue<int> q;

    q.push(Min);vis[Min]=1;cnt[Min]++;

    while(!q.empty()){

        int u=q.front();q.pop();vis[u]=0;

        for(Edge *p=e[u].next;p!=NULL;p=p->next){

            int v=p->v,w=p->w;

            if(relax(u,v,w)){

                if(!vis[v]){

                    q.push(v);

                    vis[v]=1;

                    cnt[v]++;

                    if(cnt[v]>N) return false;

                }

            }

        }

    }

    return true;

}



int main()

{

    while(cin>>N){

        memset(e,0,sizeof(e));

        Min=INF;Max=-INF;

        for(int i=0;i<N;i++){

            scanf("%d%d%d",&a,&b,&c);

            add_edge(a,b+1,c);

            if(a<Min) Min=a;

            if(b>Max) Max=b;

        }

        for(int i=Min;i<=Max;i++){

            add_edge(i,i+1,0);

            add_edge(i+1,i,-1);

        }

        spfa();

        cout<<dist[Max+1]<<endl;

    }

    return 0;

}

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