poj1716 Integer Intervals

差分约束系统构图，设S（x）为与给定的set集合的交集大小，根据相互之间的关系构图解决

code

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ls rt<<1
#define rs rt<<1|1
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 10001;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
using namespace std;
struct node{
    int s,t;
}edge[maxn];
int dis[maxn];
int main()
{
    int N,a,b,minn=INF,maxx=-INF;
    SD(N);
    FOR(i,1,N){
        SD(a);SD(b);
        edge[i].s=a;
        edge[i].t=b+1;
        minn=min(minn,a);
        maxx=max(maxx,b+1);
        //dis[i]=0;
    }
    bool flag=true;
    while(flag){
        flag=false;
        FOR(i,1,N){
            if(dis[edge[i].s]>dis[edge[i].t]-2){
                dis[edge[i].s]=dis[edge[i].t]-2;
                flag=true;
            }
        }
        FOR(i,minn,maxx-1){
            if(dis[i+1]>dis[i]+1){
                dis[i+1]=dis[i]+1;
                flag=true;
            }
        }
        FOD(i,maxx-1,minn){
            if(dis[i]>dis[i+1]){
                dis[i]=dis[i+1];
                flag=true;
            }
        }
    }
    PD(dis[maxx]-dis[minn]);LN;
    return 0;
}