poj 1201(差分约束)

poj 1201
差分约束。
设$dis[i]$为$[1,i]$包含在$Z$集合内数的个数
由题意得，$dis[b]-dis[a-1]\ge c$，
由隐含条件每个数只出现一次或不出现，得$0 \le dis[i]-dis[i-1] \le 1$
整理后得，
$$
dis[b]-dis[a-1]\ge c \\
dis[i]-dis[i-1]\ge 0 \\
dis[i-1]-dis[i]\ge -1
$$
然后按照差分约束系统用SPFA求最长路即可。

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<vector>
#include<queue>
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
const int MAXN = 50000 + 5; 
struct edge
{
    int u, v, c;
}E[MAXN*3*2];
vector<int> G[MAXN*2];
int e_num;
int n;
int minr = 100000000, maxr = -minr;
void addE(int u, int v, int c)
{
    e_num++;
    E[e_num].u = u;
    E[e_num].v = v;
    E[e_num].c = c;
    G[u].push_back(e_num);
}
int dis[MAXN*2];
int vi[MAXN*2];
void spfa()
{
    ms(dis, -(127/3));
    ms(vi, false);
    queue<int> q;
    q.push(minr); dis[minr] = 0;
    while (!q.empty())
    {
        int r = q.front(); q.pop();
        vi[r] = false;
        for (int i=0;i<G[r].size();i++)
        {
            edge ed = E[G[r][i]];
            if (dis[ed.v]<dis[r]+ed.c)
            {
                dis[ed.v] = dis[r] + ed.c;
                if (!vi[ed.v])
                {
                    vi[ed.v] = true;
                    q.push(ed.v);
                }
            }
        }
    }
}
int main()  
{  
    while (scanf("%d", &n)==1)
    {
        e_num = 0;
        for (int i=1;i<=n;i++)
        {
            int ai,bi,ci;
            scanf("%d%d%d", &ai, &bi, &ci);
            //dis[bi]-dis[ai-1]>=ci
            addE(ai-1, bi, ci);
            minr = min(minr, ai-1);
            maxr = max(maxr, bi);
        }
        for (int i=minr+1;i<=maxr;i++)
        {
            //0<=dis[i]-dis[i-1]<=1
            //dis[i]-dis[i-1]>=0
            addE(i-1, i, 0);
            //dis[i]-dis[i-1]<=1
            //dis[i-1]-dis[i]>=-1
            addE(i, i-1, -1);
        }
        spfa();
        printf("%d\n", dis[maxr]);
    }
    return 0;  
}