Intervals poj1201(差分约束)

Intervals

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25756 Accepted: 9854

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

Southwestern Europe 2002

注意！！！！原点是dis[max]  而且由于本例题s0-s11<=-M,既要求原点s11到s0的最短路径  长度为-M

假设最短路径求得各顶点到s11原点距离保存到dis[]里面  那么-M=dis[0]-dis[11],M=dis[11]-dis[0]

#include<stdio.h>#include<string.h>#include<queue>#define maxn 55000#define INF 0x3f3f3f3f#define max(a,b) a>b?a:b#define min(a,b) a>b?b:ausing namespace std;struct node{    int v,w,next;}e[maxn*3];int dis[maxn],vis[maxn],head[maxn];int cnt;void init(){    cnt=0;    memset(head,-1,sizeof(head));    memset(vis,0,sizeof(vis));}void add(int u,int v,int w){    e[cnt].v=v;    e[cnt].w=w;    e[cnt].next=head[u];    head[u]=cnt++;}void spfa(int mi,int ma){    int i;    for(i=mi;i<=ma;i++)    {         dis[i]=INF;    }    dis[ma]=0;    vis[ma]=1;    queue<int>q;    q.push(ma);    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=e[i].next)        {            int v=e[i].v;            int w=e[i].w;            if(dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;                if(!vis[v])                {                    q.push(v);                    vis[v]=1;                }            }        }    }}int main(){    int n;    int i,j;    while(~scanf("%d",&n))    {        int a,b,x;        int MIN=INF;        int MAX=-1;        init();        for(i=1;i<=n;i++)        {            scanf("%d%d%d",&a,&b,&x);            MIN=min(a,MIN);            MAX=max(b+1,MAX);            add(b+1,a,-x);        }        for(i=0;i<MAX;i++)        {            add(i+1,i,0);            add(i,i+1,1);        }        spfa(MIN,MAX);        printf("%d\n",dis[MAX]-dis[MIN]);    }}