poj1201/zoj1508/hdu1384 Intervals(差分约束)

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Intervals

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.

Write a program that:

> reads the number of intervals, their endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi
- ai + 1.

Process to the end of file.

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

题意：

对于一个序列，有n个描述，[ai,bi,ci]分别表示在区间[ai,bi]上，至少有ci个数属于该序列。输出满足这n个条件的最短的序列（即包含的数字个数最少）

分析：

设s[i]表示从0到i中有s[i]个数属于这个序列。

那么，对于每个描述，都可以得到一个方程s[bi]-s[ai-1]>=ci

同时由于每个数至多取一次，那么有s[i]-s[i-1]<=1,即s[i-1]-s[i]>=-1;

另外还有s[i]-s[i-1]>=0;

由这三个方程组建图，利用最长路即可求出。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 using namespace std;
 typedef long long ll;
 typedef pair<int,int> PII;
 #define MAXN 50010
 #define REP(A,X) for(int A=0;A<X;A++)
 #define INF  0x7fffffff
 #define CLR(A,X) memset(A,X,sizeof(A))
 struct node {
     int v,d,next;
 }edge[*MAXN];
 int head[MAXN];
 int e=;
 void init()
 {
     REP(i,MAXN)head[i]=-;
 }
 void add_edge(int u,int v,int d)
 {
     edge[e].v=v;
     edge[e].d=d;
     edge[e].next=head[u];
     head[u]=e;
     e++;
 }
 bool vis[MAXN];
 int dis[MAXN];
 void spfa(int s){
     CLR(vis,);
     REP(i,MAXN)dis[i]=i==s?:INF;
     queue<int>q;
     q.push(s);
     vis[s]=;
     while(!q.empty())
     {
         int x=q.front();
         q.pop();
         int t=head[x];
         while(t!=-)
         {
             int y=edge[t].v;
             int d=edge[t].d;
             t=edge[t].next;
             if(dis[y]>dis[x]+d)
             {
                 dis[y]=dis[x]+d;
                 if(vis[y])continue;
                 vis[y]=;
                 q.push(y);
             }
         }
         vis[x]=;
     }
 }
 int main()
 {
     ios::sync_with_stdio(false);
     int n;
     int u,v,d;
     int ans=;
     int maxn=;
     int minn=MAXN;
     while(scanf("%d",&n)!= EOF&&n)
     {
         e=;
         init();
         REP(i,n)
         {
             scanf("%d%d%d",&u,&v,&d);
             add_edge(u,v+,-d);
             maxn=max(maxn,v+);
             minn=min(u,minn);
         }
         for(int i=minn;i<maxn;i++){
             add_edge(i+,i,);
             add_edge(i,i+,);
         }
         spfa(minn);
         cout<<-dis[maxn]<<endl;
     }
     return ;
 }

代码君