POJ 1201 Intervals（差分约束系统）

Intervals

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 27491 Accepted: 10583

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

题意：有n个区间[ai,bi]，找一最短序列，要求该序列中至少有ci个数字在区间[ai,bi]上。

差分约束系统要求最小值，求出最长路径即可。根据题目给出的条件bi-ai+1>=ci，构造出来的边显然不足以求出最长路径，因为有很多点依然没有连接起来，导致可能不存在从起点到终点的路线。因此，寻找隐藏边，最大区间的点应该满足0<=点<=1。加上这个隐藏边，显然可以求出最长路径了。

代码：

#include <iostream>  #include <stack>  #include <algorithm>  #include <string.h>  using namespace std;    const int MAXN = 50005;  const int MAXE = 150005;    typedef struct Edge  {      int v, w;      int next;  }Edge;    Edge edge[MAXE];  int dist[MAXN], head[MAXN];  bool used[MAXN];  int cnt=0, n, m;int Min,Max;    void add_edge(int u, int v, int w)  {      edge[cnt].v = v;      edge[cnt].w = w;      edge[cnt].next = head[u];      head[u] = cnt++;  }    void spfa()  {      int sta[MAXN];      memset(used, false, sizeof(used));      for(int i=Min;i<=Max;i++)     dist[i] = -0x3f3f3f3f;    int top=0;      sta[++top] = Min;//进栈       dist[Min] = 0;      used[Min] = true;//标记       while (top)      {          int u = sta[top--];//出栈           used[u] = false;//去除标记           for (int i=head[u]; i!=-1; i=edge[i].next)          {              int v = edge[i].v;              int w = edge[i].w;              if (dist[v]<dist[u]+w)              {                  dist[v] = dist[u] + w;                  if (!used[v])                  {                      used[v] = true;                      sta[++top] = v;                  }              }          }      }  }  int main(){    while(~scanf("%d", &n))    {        memset(head,-1,sizeof(head));        int u,v,w;        Max=-0x3f3f3f3f,Min=0x3f3f3f3f;        for(int i=0;i<n;i++)        {            scanf("%d%d%d", &u,&v,&w);//题目条件的边            add_edge(u,v+1,w);            Min = min(Min, u);//在这里记录最小值和最大值，所求的就是以Min为源点            Max = max(Max, v+1);//以Max为终点的最短路        }        for(int i = Min;i < Max; i++)//添加新边        {            add_edge(i,i+1,0);            add_edge(i+1,i,-1);        }        spfa();        printf("%d\n", dist[Max]);    }    return 0;}