hdu 1384 查分约束

#include<stdio.h>
/*
要善于挖掘隐含条件
dis[v]-dis[u]>=bian[i].w;一个条件(u,v,bian[i].w);
dis[i+1]>=dis[i];隐含条件(i,i+1,0);
dis[i+1]-dis[i]<=1即dis[i]>=dis[i+1]-1;(i+1,i,-1);
>=求最大约束条件，求最长路.
<=求最小约束条件，求最短路.
*/
#include<string.h>
#include<queue>
using namespace std;
#define N  51000
#define inf 0x3fffffff
struct node {
int u,v,w,next;
}bian[N*4];
int head[N],yong;
void addedge(int u,int v,int w) {
bian[yong].u=u;
bian[yong].v=v;
bian[yong].w=w;
bian[yong].next=head[u];
head[u]=yong++;
}
int spfa(int s,int t) {
int dis[N];
queue<int>q;
int visit[N],i;
for(i=s;i<=t;i++)
    dis[i]=-inf;
memset(visit,0,sizeof(visit));
visit[s]=1;
dis[s]=0;
q.push(s);
while(!q.empty()) {
    int u=q.front();
    q.pop();
    visit[u]=0;
   for(i=head[u];i!=-1;i=bian[i].next) {
    int v=bian[i].v;
    if(dis[v]<dis[u]+bian[i].w) {
        dis[v]=dis[u]+bian[i].w;
        if(!visit[v]) {
            visit[v]=1;
            q.push(v);
        }
    }
   }
}
return dis[t];
}
int main()
{
    int n,i,j,k,minn,maxx;
    while(scanf("%d",&n)!=EOF)
    {
        minn=inf;
        maxx=-1;
        memset(head,-1,sizeof(head));
        yong=0;
        while(n--)
        {
            scanf("%d%d%d",&i,&j,&k);
            addedge(i,j+1,k);
            if(j+1>maxx)maxx=j+1;
            if(i<minn)minn=i;
        }
        for(i=minn; i<=maxx; i++)
           addedge(i,i+1,0),addedge(i+1,i,-1);//
       printf("%d
",spfa(minn,maxx));
        }
    return 0;
}