Intervals

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

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

这个题开始用的简单队列，结果超时了，然后就用了数组来存。而且开始拉掉了一个条件s[i+1]-s[i]<=1

这个题有：0<=s[i+1]-s[i]<=1(s[i]-s[i+1]>=-1)  s[b]-s[a]>=w

因为是求最小值所以用>=

#include <iostream>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#define oo 1<<28#include<queue>struct node{    int u;    int v;    int w;    int next;} edge[1100000];int n,m,s,t;using namespace std;int cnt;int dis[51000];int head[51000];int vist[51000];int pre[51000];int num[51000];void add(int u,int v,int w){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;}void init(){    for(int i=0; i<=n+1; i++)    {        dis[i]=-oo;    }    memset(pre,-1,sizeof(pre));    memset(vist,0,sizeof(vist));    memset(num,0,sizeof(num));}void Spfa(){    int i,cou=0;    init();    int q[35000];//用单纯的队列会超时    dis[s]=0;    vist[s]=1;    q[cou++]=s;    while(cou)    {        cou--;        int t=q[cou];        i=head[t];        vist[t]=0;        while(i!=-1)        {            int v=edge[i].v;            int u=edge[i].u;            int w=edge[i].w;            if(dis[v]<w+dis[t])            {                dis[v]=w+dis[t];                if(!vist[v])                {                    vist[v]=1;                    q[cou++]=v;                }            }            i=edge[i].next;        }    }    printf("%d\n",dis[t]-dis[s]);}int main(){    int i,j;    char str[1000];    while(~scanf("%d",&n))    {        int u,v,w;        int maxx=0,minn=oo;        cnt=0;        memset(head,-1,sizeof(head));        for(i=1; i<=n; i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v+1,w);//s[v+1]-[u]>=w求最小值用>=            minn=min(u,minn);            maxx=max(v+1,maxx);        }        s=minn,t=maxx;        for(i=minn;i<maxx;i++)        {            add(i,i+1,0);//s[i+1]-s[i]>=0            add(i+1,i,-1);//开始拉掉了这个条件然后看的题解发现s[i+1]-s[i]<=1--->s[i]-s[i+1]>=-1        }        Spfa();    }    return 0;}