POJ 1201 Intervals（差分约束）

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=200010,inf=~0U>>2,M=200000;
int time[N],q[N],size,h,t,n,ed,dis[N],in[N],nxt[N],w[N],v[N],g[N],ma,mi,u,e,cost;
void add(int x,int y,int z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
int spfa(int S){
    for(int i=0;i<=n;i++)dis[i]=-inf,in[i]=0,time[i]=0;
    time[S]=1,dis[S]=0,in[S]=1;
    int i,x,size; q[h=t=size=1]=S;
    while(size){
        for(i=g[x=q[h]],h=(h+1)%M,size--;i!=-1;i=nxt[i])if(dis[x]+w[i]>dis[v[i]]){
            dis[v[i]]=dis[x]+w[i];
            if(!in[v[i]]){
                time[v[i]]++,t=(t+1)%M,size++,in[q[t]=v[i]]=1;
                if(time[v[i]]>n)return -1;
            }
        }in[x]=0;
    }if(dis[n]==-inf)return -2;
    return dis[n];
}
int main(){
    while(~scanf("%d",&n)){
        ed=0;memset(g,-1,sizeof(g));
        ma=-inf; mi=inf;
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&u,&e,&cost);
            add(u,e+1,cost);
            ma=max(e+1,ma); mi=min(u,mi);
        }for(int i=mi;i<ma;i++)add(i,i+1,0),add(i+1,i,-1); 
        n=ma; printf("%d\n",spfa(mi));
    }return 0;
}