Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18740		Accepted: 7042

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

Source

 

 

#include 
      
       #include 
       
        #define EM 300000#define VM 50005struct edge{    int v,w,next;}e[EM];int head[VM],ep;void addedge(int cu,int cv,int cw){    ep ++;    e[ep].v = cv;    e[ep].w = cw;    e[ep].next = head[cu];    head[cu] = ep;}int maxn (int a,int b){    return a > b?a:b;}void spfa (int n){    int vis[VM],dis[VM],stack[EM];    memset (vis,0,sizeof(vis));    memset (dis,-1,sizeof(dis));    //for (int i = 0;i <= n;i ++)     //   dis[i] = inf;    dis[n+2] = 0;    int top = 1;    vis[n+2] = 1;    stack[0] = n+2;    while (top)    {        int u = stack[--top];        vis[u] = 0;        for (int i = head[u];i != -1;i = e[i].next)        {            int v = e[i].v;            if (dis[v] < dis[u] + e[i].w)            {                dis[v] = dis[u] + e[i].w;                if (!vis[v])                {                    vis[v] = 1;                    stack[top++] = v;                }            }        }    }    printf ("%d\n",dis[n]);}int main (){    int n,v1,v2,m,cost;    ep = 0;    scanf ("%d",&n);    m = n;    memset (head,-1,sizeof(head));    int max = -1;    while (m --)    {        scanf ("%d%d%d",&v1,&v2,&cost);        addedge (v1,v2+1,cost);        max = maxn (max,v2+1);    }    for (int i = 0;i < max;i ++)    {        addedge (i,i+1,0);        addedge (i+1,i,-1);        addedge (max+2,i,0);    }    spfa (max);    return 0;}