HDU 1384 Intervals（差分约束）

problem

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

在 [ai,bi] [ a i , b i ] 至少有 ci c i 个相同的数，求满足条件的最小集合大小

题解

求最小值，即对约束图求最长路，对于线性约束条件： dj−di≥x d j − d i ≥ x ，则添加边 (i,j) ( i , j ) 且 w(i,j)=x w ( i , j ) = x
题意中的条件为： d[bi]−d[ai−1]≥ci d [ b i ] − d [ a i − 1 ] ≥ c i 即加边 (ai−1,bi) ( a i − 1 , b i ) 且 w(ai−1,bi)=ci w ( a i − 1 , b i ) = c i
题中有隐藏条件： 0≤d[i+1]−d[i]≤1 0 ≤ d [ i + 1 ] − d [ i ] ≤ 1
0≤d[i+1]−d[i] 0 ≤ d [ i + 1 ] − d [ i ] 即加边 (i,i+1) ( i , i + 1 ) 且 w(i,i+1)=0 w ( i , i + 1 ) = 0
d[i+1]−d[i]≤1 d [ i + 1 ] − d [ i ] ≤ 1 进行等价变换，得 d[i]−d[i+1]≥−1 d [ i ] − d [ i + 1 ] ≥ − 1 即加边 (i+1,i) ( i + 1 , i ) 且 w(i+1,i)=−1 w ( i + 1 , i ) = − 1

代码

#include
using namespace std;

const int maxn = 5e4+100;

struct node
{
    int to,w;
    node (int n1,int n2)
    {
        to = n1;
        w = n2;
    }
};
vector  v[maxn];
int n,le,ri,l,r,c;
int d[maxn];
bool vis[maxn];

void spfa()
{
    for (int i=le;i<=ri;i++)
    {
        d[i] = -maxn;
        vis[i] = 0;
    }
    queue  Q;
    Q.push(le);
    vis[le] = 1;
    d[le] = 0;
    while (!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for (int i=0;iif (d[t] < d[u]+dt)
            {
                d[t] = d[u]+dt;
                if (!vis[t])
                {
                    vis[t] = true;
                    Q.push(t);
                }
            }
        }
    }
    return;
}

int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        for (int i=0;i0,le = maxn;
        for (int i=0;i"%d %d %d",&l,&r,&c);
            l++,r++;
            le = min(l,le);
            ri = max(r,ri);
            v[l-1].push_back(node(r,c));
        }
        le--;
        for (int i=le;i1,0));
            v[i+1].push_back(node(i,-1));
        }

        spfa();
        printf("%d\n",d[ri]);
    }
    return 0;
}