/*题意是给定区间[ai,bi]，要就在区间上至少有ci个点，如果用S[i+1]表示区间[0，i]的点数,则题目要求就可以转化为S[bi+1]-S[ai]>=ci;联系到求单源最短路时的松弛操作：若dis[v]>dis[u]+w[u][v]，则dis[v]=dis[u]+w[u][v] ；即要求dis[v]<=dis[u]+w[u][v],对比此处的S[bi+1]>=ci+S[ai];所以本题可以用求最长路的方法求解。当然也可以按照差分约束的理论转为求最短路，都差不多。代码一：转换为最短路问题*/#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<map>#include<algorithm>using namespace std;const int maxn=50002;const int MAXM=1000009;const int inf=999999999;int n,m;struct Edge{int v,w;int next;}edge[4*maxn];int e,top;int first[maxn];int vis[maxn];int Stack[10*maxn];void addedge(int u,int v,int w){edge[e].v=v;edge[e].w=w;edge[e].next=first[u];first[u]=e++;}int maxpoint;int minpoint;int d[maxn];void spfa(){for(int i=minpoint;i<=maxpoint;i++){addedge(i,i+1,1);addedge(i+1,i,0);}for(int i=minpoint;i<=maxpoint;i++)d[i]=inf;d[maxpoint]=0;int front=-1,rear=-1;Stack[++rear]=maxpoint;while(front<rear){int x=Stack[(++front)%maxn];vis[x]=0;for(int k=first[x];k!=-1;k=edge[k].next){//cout<<"a"<<endl;if(d[edge[k].v]>d[x]+edge[k].w){d[edge[k].v]=d[x]+edge[k].w;if(!vis[edge[k].v]){vis[edge[k].v]=1;Stack[(++rear)%maxn]=edge[k].v;}}}// for(int i=0;i<=maxpoint;i++)// cout<<i<<" "<<d[i]<<endl;}printf("%d\n",-d[minpoint]);}int main(){while(scanf("%d",&n)!=EOF){e=0;maxpoint=-1;minpoint=inf;top=-1;int a,b,c;memset(vis,0,sizeof(vis));memset(first,-1,sizeof(first));for(int i=0;i<n;i++){scanf("%d%d%d",&a,&b,&c);a++,b++;addedge(b,a-1,-c);if(b>maxpoint)maxpoint=b;if(a-1<minpoint)minpoint=a-1;}spfa();}}// 代码二：转换为最长路问题#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<map>#include<algorithm>using namespace std;const int maxn=50002;const int MAXM=1000009;const int inf=999999999;int n,m;struct Edge{int v,w;int next;}edge[4*maxn];int e,top;int first[maxn];int vis[maxn];int Stack[10*maxn];void addedge(int u,int v,int w){edge[e].v=v;edge[e].w=w;edge[e].next=first[u];first[u]=e++;}int maxpoint;int minpoint;int d[maxn];void spfa(){for(int i=minpoint;i<=maxpoint;i++){addedge(i,i+1,0);addedge(i+1,i,-1); //建边方向不同} for(int i=minpoint;i<=maxpoint;i++)d[i]=-inf; //初始化值不同d[minpoint]=0; //起点不同int front=-1,rear=-1;Stack[++rear]=minpoint;while(front<rear){int x=Stack[(++front)%maxn];vis[x]=0;for(int k=first[x];k!=-1;k=edge[k].next){//cout<<"a"<<endl;if(d[edge[k].v]<d[x]+edge[k].w) //松弛条件不同{d[edge[k].v]=d[x]+edge[k].w;if(!vis[edge[k].v]){vis[edge[k].v]=1;Stack[(++rear)%maxn]=edge[k].v;}}}// for(int i=0;i<=maxpoint;i++)// cout<<i<<" "<<d[i]<<endl;}printf("%d\n",d[maxpoint]-d[minpoint]);}int main(){while(scanf("%d",&n)!=EOF){e=0;maxpoint=-1;minpoint=inf;top=-1;int a,b,c;memset(vis,0,sizeof(vis));memset(first,-1,sizeof(first));for(int i=0;i<n;i++){scanf("%d%d%d",&a,&b,&c);a++,b++;addedge(a-1,b,c); //建边方向不一样if(b>maxpoint)maxpoint=b;if(a-1<minpoint)minpoint=a-1;}spfa();}}
