poj1201——Intervals（差分约束）

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

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
Sample Output

6

给出一些区间[a,b]和对应的数字c，表示要在区间内至少选c个数，求一共最少能选多少个数？
要转化成差分约束系统，设sa是[0,a]选出的数，则sb-sa<=c。
另外0<=S(i+1)-Si<=1，即i+1这个点取了或者没取，这样就有三条边了

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <math.h>#include <algorithm>#include <queue>#include <iomanip>#include <map>#include <cctype>#define INF 0x3f3f3f3f#define MAXN 500005#define Mod 1000000007using namespace std;struct Edge{    int v,w,next;};Edge edge1[MAXN<<1];int head1[MAXN],n,m,e,vis[MAXN],dis[MAXN];int s,t;void add(Edge *edge,int *head,int u,int v,int w){    edge[e].v=v;    edge[e].w=w;    edge[e].next=head[u];    head[u]=e;    e++;}void spfa(Edge *edge,int *head,int u){    memset(vis,0,sizeof(vis));    for(int i=s;i<=t;++i)        dis[i]=-INF;    dis[u]=0;    queue<int> q;    q.push(u);    while(!q.empty())    {        u=q.front();        q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v,w=edge[i].w;            if(w+dis[u]>dis[v])            {                dis[v]=w+dis[u];                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }}int main(){    while(~scanf("%d",&m))    {        e=0;        memset(head1,-1,sizeof(head1));        s=INF;        t=-INF;        for(int i=0;i<m;++i)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            add(edge1,head1,a,b+1,c); //防止区间重叠            s=min(s,a);            t=max(t,b+1);        }        for(int i=s;i<t;++i)        {            add(edge1,head1,i,i+1,0);            add(edge1,head1,i+1,i,-1);        }        spfa(edge1,head1,s);        printf("%d\n",dis[t]);    }    return 0;}