POJ1201（Hdu1384） Intervals差分约束系统

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

题意大概是这样：有n个区间[Ai，Bi],有一个集合Z，集合Z和每个区间至少有Ci个相同元素；求这个集合至少有多少个元素。
定义dis[n]数组记录区间[0,n]的最少有多少个元素在集合z内，则可以建立一个不等式：dis[Bi+1]-dis[Ai]>=Ci；
另外由题意有：0<=dis[i+1]-dis[i]<=1;（因为i可能从0开始所以不使用i-1和i）
因此有：
dis[i+1]-dis[i]>=0；
dis[i]-dis[i+1]>=-1；
因为是求最小值，所以是计算最长路。
边数比较多而且有负边所以使用spfa算法。

代码如下：

#include
#include
#include
#include
#include
#define INF 1<<29
using namespace std;
struct node
{
    int to,len,next;
}edge[200050];
int head[50010],used[50010],dis[50010],inqueue[50010],k,n,minx,maxx;
void add_edge(int st,int en,int len)
{
    edge[k].to=en;
    edge[k].len=len;
    edge[k].next=head[st];
    head[st]=k++;
}
void init()
{
    memset(used,0,sizeof(used));
    memset(inqueue,0,sizeof(inqueue));
    memset(head,-1,sizeof(head));
    minx=INF,maxx=-1;
    for(int i=1;i<=n;i++)
    {
        int st,en,len;
        scanf("%d%d%d",&st,&en,&len);
        minx=min(minx,st);
        maxx=max(maxx,en+1);
        add_edge(st,en+1,len);
    }
    for(int i=minx;i<=maxx;i++)
    {
        add_edge(i,i+1,0);
        add_edge(i+1,i,-1);
    }
}
void spfa(int st,int time)
{
    for(int i=minx;i<=maxx;i++)
        dis[i]=-INF;
    queue<int>q;
    dis[st]=0;
    used[st]=1;
    q.push(st);
    inqueue[st]++;
    while(!q.empty())
    {
        int top=q.front();
        q.pop();
        used[top]=0;
        for(int j=head[top];j!=-1;j=edge[j].next)
        {
            if(dis[edge[j].to]if(!used[edge[j].to])
                {
                    used[edge[j].to]=1;
                    if(++inqueue[edge[j].to]>time-1)
                    {
                        printf("%d\n",dis[maxx]);
                    }
                   q.push(edge[j].to);
                }
            }
        }
    }
    printf("%d\n",dis[maxx]);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        k=0;
        init();
        spfa(minx,maxx-minx+1);
    }
    return 0;
}