【POJ1201】Intervals

题目链接

Intervals

题目描述

You are given \(n\) closed, integer intervals \([a_i,b_i]\) and \(n\) integers \(c_1\), ...,\(c_n\).
Write a program that:
reads the number of intervals, their end points and integers \(c_1\), ...,\(c_n\) from the standard input,
computes the minimal size of a set \(Z\) of integers which has at least \(c_i\) common elements with interval \([a_i,b_i]\), for each \(i=1,2\),...,\(n\),
writes the answer to the standard output.

输入格式

The first line of the input contains an integer \(n\) (\(1 \le n \le 50000\)) -- the number of intervals. The following n lines describe the intervals. The (\(i+1\))-th line of the input contains three integers \(a_i\), \(b_i\) and \(c_i\) separated by single spaces and such that \(0 \le a_i \le b_i \le 50000\) and \(1 \le c_i \le b_i-a_i+1\).

输出格式

The output contains exactly one integer equal to the minimal size of set \(Z\) sharing at least \(c_i\) elements with interval \([a_i, b_i]\), for each \(i=1,2\),...,\(n\).

样例输入

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

样例输出

6

题解

题意：给定\(n\)个条件，要求区间\([a_i,b_i]\)之间至少有\(c_i\)个元素，\(0 \le c_i \le b_i-a_i+1\)，求最少有多少个元素。
我们令\(s[i]\)表示小于等于\(i\)的元素有多少个。
那么对于每个约束条件，就有\(s[b_i]-s[a_i-1] \ge c_i\)。
也就是\(s[b_i]+\)(\(-c_i\))\(\ge s[a_i-1]\)
这个式子和最短路的式子是一样的，所以我们可以建一条\(s[b_i]\)到\(s[a_i-1]\)长度为\(c_i\)的边。
但是我们发现这张图是不完整的。
所以我们还要加上一些隐藏条件。

• \(s[i]-s[i-1] \ge 0\)
  \(\rightarrow s[i]+0 \ge s[i-1]\)
• \(s[i]-s[i-1] \le 1\)
  \(\rightarrow s[i-1]+1 \ge s[i]\)
  所以我们加上\(s[i]\)到\(s[i-1]\)长度为\(0\)的边，
  \(s[i-1]\)到\(s[i]\)长度为\(1\)的边。
  假设最大值为\(mx\)，接下来来我们跑\(s[mx]\)到\(s[0]\)的最短路，答案就是\(s[mx]-s[0]\)。
  上代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int x,y,c;
struct aa{
    int to,nxt,v;
}p[500009];
int h[50009],len;
int mx;
int s[50009];
void add(int u,int v,int w){
    p[++len].to=v;
    p[len].v=w;
    p[len].nxt=h[u];
    h[u]=len;
}
bool k[50009];
void dfs(int u){
    int q[500009],l=0,r=0;
    q[0]=u;
    while(l!=r+1){
        u=q[l];
        l++;l%=500000;
        k[u]=0;
        for(int j=h[u];j;j=p[j].nxt){
            if(s[p[j].to]>s[u]+p[j].v){
                s[p[j].to]=s[u]+p[j].v;
                r++;
                r%=500000;//循环队列
                if(!k[p[j].to]) q[r]=p[j].to;
                k[p[j].to]=1;
            }
        }
    }
}
int main(){
    while(scanf("%d",&n)!=EOF){
        memset(s,1,sizeof(s));
        memset(h,0,sizeof(h));
        len=0;
        for(int j=1;j<=n;j++){
            scanf("%d%d%d",&x,&y,&c);
            x++;y++;
            mx=max(mx,y);
            add(y,x-1,-c);
        }
        for(int j=1;j<=mx;j++){
            add(j-1,j,1);
            add(j,j-1,0);
        }
        s[mx]=0;
        dfs(mx);
        printf("%d\n",-s[0]);
    }
    return 0;
}