POJ1201 Intervals[差分约束系统]

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26028		Accepted: 9952

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

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题意：有n个区间，每个区间有3个值，ai，bi，ci代表，在区间[ai,bi]上至少要选择ci个整数点，ci可以在区间内任意取不重复的点
现在要满足所有区间的自身条件，问最少选多少个点

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一段区间，想到前缀和处理选的个数

每个前缀和建一个点

s[b]-s[a-1]>=c

同时要满足前缀和的性质，即：

s[i]-s[i-1]>=0

s[i]-s[i-1]<=1

最小值，按>=建图跑最长路

 

注意区间范围0..n，读入时改成了1...n+1(n)

但是0节点也是（前缀和！）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=5e4+5,M=15e4+5,INF=1e9;
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;
}
int n,m,a,b,c;
struct edge{
    int v,ne;
    double w;
}e[M];
int h[N],cnt=0;
inline void ins(int u,int v,int w){
    cnt++;
    e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
}
int q[N],head,tail,inq[N],num[N],d[N];
inline void lop(int &x){if(x==N) x=1;else if(x==0) x=N-1;}
bool spfa(){
    head=tail=1;
    memset(inq,0,sizeof(inq));
    memset(num,0,sizeof(num));
    for(int i=0;i<=n;i++) q[tail++]=i,inq[i]=1,d[i]=0;
    while(head!=tail){
        int u=q[head++];inq[u]=0;lop(head);
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v,w=e[i].w;
            if(d[v]<d[u]+w){
                d[v]=d[u]+w;
                if(!inq[v]){
                    inq[v]=1;
                    if(++num[v]>n) return true;
                    if(d[v]>d[q[head]]) head--,lop(head),q[head]=v;
                    else q[tail++]=v,lop(tail);
                }
            }
        }
    }
    return false;
}
int main(){
    m=read();
    for(int i=1;i<=m;i++){
        a=read()+1;b=read()+1;c=read();n=max(n,b);
        ins(a-1,b,c);
    }
    for(int i=1;i<=n;i++) ins(i-1,i,0),ins(i,i-1,-1);
    spfa();
    printf("%d",d[n]);
}