HDU 1384 Intervals（差分约束）

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4292    Accepted Submission(s): 1624

Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

 

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

 

Author

1384

 

需要求S[r]-S[l-1]>=ans，即S[l-1]-S[r]<=-ans。若以r为起点 ，而-ans就为r到l-1的最短路径，即-dist[Maxl-1]。

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define MAXN 50100
using namespace std;

struct Edge{
    int to,nxt,w;
}e[MAXN<<2];
int dis[MAXN],head[MAXN];
bool vis[MAXN];
int cnt,n,l,r;
queue<int>q;

void init()
{
    memset(head,0,sizeof(head));
    cnt = 0;
    l = 50100;
    r = 0;
}
void add(int u,int v,int w)
{
    ++cnt;
    e[cnt].w = w;
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
}
void spfa()
{
    memset(dis,0x3f,sizeof(dis));
    memset(vis,false,sizeof(vis));
    q.push(r);
    vis[r] = true;
    dis[r] = 0;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i=head[u]; i; i=e[i].nxt)
        {
            int v = e[i].to;
            int w = e[i].w;
            if (dis[v]>dis[u]+w)
            {
                dis[v] = dis[u]+w;
                if (!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
        vis[u] = false ;
    }
    printf("%d
",-dis[l-1]);
}
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        init()
        for (int a,b,c,i=1; i<=n; ++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            l = min(l,a);
            r = max(r,b);
            add(b,a-1,-c);    //b-(a-1)>=c
        }
        for (int i=l; i<=r; ++i)
        {
            add(i,i-1,0);    //i-(i-1)>=0
            add(i-1,i,1);    //i-(i-1)<=1
        }
        spfa(); 
    } 
    return 0;
}