poj 1201

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

题意：

给你n个闭区间

【a，b】  及c

要求区间内的点数必须要大于等于c

点只存在于区间内。

然后问负无穷到正无穷最多有多少点。

做法：

我们从最左边到最右边。

区间内最少有c个点，说明点(a-1)到b点 最少会增加c个点。

也就是说  b-(a-1)>=c  转换下   (a-1)-b<=-c 就是差分约束的公式了。

然后就是建边 建一条权值为-c   的b到(a-1)的边就行了。

然后取最小的坐标x，取最大的坐标d。

因为求最大值  所以得公式 d-(x-1)>=?

加个负号， (x-1)-d<=-?   ,所以最后最短路 计算 d点到(x-1)的最短距离，然后取负就是答案了。

但是这样条件还不够。

还有关键的是相邻点之间的建边，设相邻两点  (i) 和(i+1)   

走到(i+1)数到的点数  肯定是比左边的(i)要大的 ， 所以 (i+1)-(i)>=0  加个负号， (i)-(i+1)<=0

而从(i)走到(i+1)最多只增加了一个点，所以(i+1)-(i)<=1 ，   

所以还要建这些边，才能跑出最后的答案。

//china no.1#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>using namespace std;#define pi acos(-1)#define endl '\n'#define rand() srand(time(0));#define me(x) memset(x,-1,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0);typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,-1,-1,1,1};const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=1e5+5;const int maxx=1e5+100;const double EPS=1e-7;const int MOD=10000007;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W whileint n,x,y,k;struct node{    int to,w,next;}E[maxx*40];int head[maxx];int d[maxx],inq[maxx],num[maxx];inline int Scan(){    int res=0,ch,flag=0;    if((ch=getchar())=='-')flag=1;    else if(ch>='0' && ch<='9')res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';    return flag ? -res : res;}void init(){    memset(head,-1,sizeof(head));    k=0;    for(int i=0;i<maxx;i++)    {        inq[i]=0;        d[i]=INF;    }}void add(int from,int to,int w){    E[k].to=to;    E[k].w=w;    E[k].next=head[from];    //cout<<E[k].to<<"  "<<E[k].w<<endl;    head[from]=k++;}int SPFA(){    int s=y;    queue<int >Q;    Q.push(s),d[s]=0,inq[s]=1;    me(num);    while(!Q.empty())    {        int now=Q.front();        //cout<<now<<endl;        Q.pop();        inq[now]=0;        ++num[now];        if(num[now]>n) return 0;        for(int i=head[now];i!=-1;i=E[i].next)        {            int v=E[i].to,w=E[i].w;            if(d[v]>d[now]+w)            {                d[v]=d[now]+w;                if(inq[v]==1) continue;                inq[v]=1;                Q.push(v);            }        }    }    if(d[x-1]!=INF)        cout<<-d[x-1]<<endl;    else  puts("-2");    return 1;}int main(){    int T;    cin>>T;    init();    x=INF;y=0;    W(T--)    {        int a,b,c;        a=Scan();b=Scan();c=Scan();        add(b,a-1,-c); //b到a-1        //add(b,a-1,-c);        x=min(x,a,b);        y=max(y,a,b);    }    n=y-(x-2);    FOr(x-1,y,i)    {        add(i,i+1,1);        add(i+1,i,0);        // E[i].push_back(make_pair(i+1,1));        // E[i+1].push_back(make_pair(i,0));    }    //cout<<k<<endl;    int t=SPFA();    if(t==0)        puts("-1");}