poj 1201 Intervals

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21863		Accepted: 8241

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

关键是怎么构造差分约束模型啊，这题设s[i]表示集合Z中小于等于i的个数，那么集合Z中[ai,bi]的个数就是s[bi]-s[ai-1],考虑到题目中ai最小为0，而s[ai-1]就越界了，因此将输入同步+1；

我们可以得到s[bi]-s[ai-1]>=ci,也就是s[ai-1]-s[bi]<=-ci,同时s[i]-s[i-1]<=1&&s[i-1]-s[i]<=0,之后就是简单的构造有向图了，我们来看看需要求的量，Z中元素的最小个数。

我们用变量maxx记录区间最右端的值，minx记录区间最左端的值，那么就是求|Z|=S[maxx]-S[minx-1]的最小值。假设S[maxx]-S[minx-1]>=M,则当M取最大时，左侧达到最小值。变型后S[minx-1]-S[maxx]<=-M,

M取大，也就是-M取小，也就是maxx->minx-1达到最短路径，M就是最短路径的相反数了。因此以maxx为源点，一遍最短路即可。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#define Maxn 50010
using namespace std;

struct line{
    int v,w;
    line(int vv=0,int ww=0):v(vv),w(ww){}
}p[Maxn*3];
int head[Maxn],dist[Maxn],vis[Maxn],nxt[Maxn*3];
int q[Maxn];
int tot,l,r;
const int inf=0x3f3f3f3f;
void addedge(int u,int v,int w){
    p[tot]=line(v,w);
    nxt[tot]=head[u];
    head[u]=tot++;
}
void spfa(int u){
    for(int i=l-1;i<=r;i++)
        dist[i]=inf,vis[i]=0;
    dist[u]=0;
    int s=0,e;
    q[e=1]=u;
    while(s!=e){
        int v=q[s=(s+1)%Maxn];
        vis[v]=0;
        for(int i=head[v];i!=-1;i=nxt[i]){
            int r=p[i].v,w=p[i].w;
            if(dist[v]+w<dist[r]){
                dist[r]=dist[v]+w;
                if(!vis[r]){
                    vis[r]=1;
                    q[e=(e+1)%Maxn]=r;
                }
            }
        }
    }
}
int main()
{
    int n,fr,to,w;
    while(~scanf("%d",&n)){
        memset(head,-1,sizeof head);
        tot=0;
        l=inf,r=0;
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&fr,&to,&w);
            addedge(to+1,fr,-w);
            l=min(l,fr+1);
            r=max(r,to+1);
        }
        for(int i=l;i<=r;i++){
            addedge(i-1,i,1);
            addedge(i,i-1,0);
        }
        spfa(r);
        printf("%d\n",-dist[l-1]);
    }
	return 0;
}