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;
}