POJ1201 Intervals

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

正解：SPFA+差分约束系统

解题报告：

　　大致题意是给定一些要求，比如说，1到7之间至少有5个数，然后条件需要全部满足，问

　　这几天刷了几道差分约束系统+SPFA的题目，现在向总给的题目清单里面就只剩下一道半平面交没做了。夏令营回来有时间再切了吧。

　　这道题也没什么好说的，只不过是根据大于号建图，跑最长路就可以了。

　　唯一要注意的是因为是前缀和建图，需要把这个序列依次连边，比如说1连2连3这么一直连接下去。

 //It is made by jump~
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <ctime>
 #include <vector>
 #include <queue>
 #include <map>
 #ifdef WIN32
 #define OT "%I64d"
 #else
 #define OT "%lld"
 #endif
 using namespace std;
 typedef long long LL;
 const int inf = (<<);
 const int MAXN = ;
 const int MAXM = ;
 int n;
 int first[MAXN],next[MAXM],to[MAXM],w[MAXM];
 int dis[MAXN];
 bool pd[MAXN];
 int ecnt;
 int Min,Max;
 int ans;
 queue<int>Q;

 inline int getint()
 {
        int w=,q=;
        char c=getchar();
        while((c<'' || c>'') && c!='-') c=getchar();
        if (c=='-')  q=, c=getchar();
        while (c>='' && c<='') w=w*+c-'', c=getchar();
        return q ? -w : w;
 }

 inline void Init(){
     ecnt=; memset(first,,sizeof(first));
     Max=; Min=inf;
     while(!Q.empty()) Q.pop();
     ans=;
 }

 inline void link(int x,int y,int z){
     next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=y; w[ecnt]=z;
 }

 inline void spfa(){
     for(int i=;i<=n;i++) dis[i]=-inf;
     Q.push(Min); pd[Min]=; dis[Min]=;
     while(!Q.empty()) {
     int u=Q.front(); Q.pop(); pd[u]=;
     for(int i=first[u];i;i=next[i]) { //最长路
         int v=to[i];
         if(dis[v]<dis[u]+w[i]) {
         dis[v]=dis[u]+w[i];
         if(!pd[v]){
             Q.push(v); pd[v]=;
         }
         }
     }
     }
     ans=dis[Max];
 }

 inline void solve(){
     while(scanf("%d",&n)!=EOF){
         Init();
     int x,y,z;
     for(int i=;i<=n;i++) {
         x=getint()-;y=getint();z=getint();
         link(x,y,z);
         Min=min(Min,x); Max=max(Max,y);
     }
     for(int i=Min;i<Max;i++){//以前缀和为结点
         link(i+,i,-); link(i,i+,);//要使所有点满足s[i+1]-s[i] <= 1 &&  s[i]-s[i+1] <= 0
     }
     spfa();
     printf("%d",ans);
     }
 }

 int main()
 {
   solve();
   return ;
 }