Intervals（poj1201

                         题目传送门

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28676		Accepted: 11065

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

Source

Southwestern Europe 2002

 

　大意：给你n个区间，每个区间[li,ri]至少选出ci个数，问最小可行集合的大小。

　 题解：

　　一道典型的差分约束问题。（没学过差分约束的同学可以看这篇讲解：传送门）

　　我们设d[i]表示前i个数中被选的数的个数，则区间限制变成了d[ri]-d[li-1]>=ci. 建图之后设跑最长路即可。

　　需要注意的是题目中的隐含条件：d[i]>=d[i-1],d[i-1]>=d[i]-1

　　又由于[0,x]这样的区间会出现d[-1]这样的情况，所以我们稍作改变，让d[i]表示前i-1个数中被选数的个数，起点为d[0]=0.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<bitset>
#define LL long long
#define RI register int
using namespace std;
const int INF = 0x7ffffff ;
const int N = 50000 + 10 ;

inline int read() {
    int k = 0 , f = 1 ; char c = getchar() ;
    for( ; !isdigit(c) ; c = getchar())
      if(c == '-') f = -1 ;
    for( ; isdigit(c) ; c = getchar())
      k = k*10 + c-'0' ;
    return k*f ;
}
struct Edge {
    int to, next, val ;
}e[N<<2] ;
int n ; int head[N], dis[N], num[N] ;
inline void add_edge(int x,int y,int z) {
    static int cnt = 0 ;
    e[++cnt].to = y, e[cnt].next = head[x], head[x] = cnt, e[cnt].val = z ;
}

inline bool spfa() {
    for(int i=0;i<N;i++) dis[i] = INF ;
    queue<int>q ; q.push(0), dis[0] = 0 ; bitset<N>inq ; inq[0] = 1 ;
    while(!q.empty()) {
        int x = q.front() ; q.pop() ;
        if(num[x] > n) return 0 ;
        for(int i=head[x];i;i=e[i].next) {
            int y = e[i].to ; 
            if(dis[y] > dis[x]+e[i].val) {
                dis[y] = dis[x]+e[i].val ;
                if(!inq[y]) {
                    q.push(y) ; inq[y] = 1 ; num[y]++ ;
                }
            }
        }
        inq[x] = 0 ;
    }
    return 1 ;
}

int main() {
    n = read() ; int maxx = 0, minn = INF ;
    for(int i=1;i<=n;i++) {
        int x = read(), y = read(), z = read() ;
        maxx = max(maxx,y+1), minn = min(minn,x) ;
        add_edge(x,y+1,-z) ;
    }
    add_edge(0,minn,0) ;
    for(int i=minn;i<=maxx;i++) add_edge(i,i+1,0), add_edge(i+1,i,1) ;
    if(!spfa()) {
        printf("No solution!") ; return 0 ;
    } 
    printf("%d",-dis[maxx]) ;
    return 0 ;
}