POJ 1201 Intervals

d[i]=区间[0,i)中所有的整点数。得到两个不等式：

d[b+1]-d[a]<=c,0<=d[i+1]-d[i]<=1;

根据这两个不等式来做差分约束。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 50010
#define maxm 150010
using namespace std;

int first[maxn],d[maxn],inq[maxn];
int v[maxm],next[maxm],w[maxm];
int e,n;

void init()
{
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,int c)
{
    v[e] = b,next[e] = first[a],first[a] = e,w[e] = c,e++;
}

int spfa(int min1,int max1)
{
    queue<int> q;
    int i,ith;
    for(i = min1;i <= max1;i++)
    {
        d[i] = -maxn;
        inq[i] = 0;
    }
    d[min1] = 0;
    q.push(min1);
    inq[min1] = 1;
    while(!q.empty())
    {
        ith = q.front();
        q.pop();
        inq[ith] = 0;
        for(i = first[ith];i != -1;i = next[i])
        {
            if(d[v[i]] < d[ith] + w[i])
            {
                d[v[i]] = d[ith] + w[i];
                if(!inq[v[i]])
                {
                    q.push(v[i]);
                    inq[v[i]] = 1;
                }
            }
        }
    }
    return d[max1];
}
int main()
{
    while(scanf("%d",&n) == 1)
    {
        int max1 = -maxn,min1 = maxn;
        int i,a,b,c;
        init();
        for(i = 1;i <= n;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add_edge(a,b + 1,c);
            max1 = b > max1 ? b : max1;
            min1 = a < min1 ? a : min1;
        }
        max1++;
        for(i = min1;i < max1;i++)
        {
            add_edge(i,i + 1,0);
            add_edge(i + 1,i,-1);
        }
        printf("%d\n",spfa(min1,max1));
    }
    return 0;
}