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Description
During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.
snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?
题目就是差分约束问题，但是这个题卡SPFA。。。。。。
代码如下：
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MaxN=30004; const int MaxM=150004; const int INF=10e9+7; struct Node { int id,val; Node(int _id=0,int _val=0):id(_id),val(_val) {} bool operator < (const Node & a) const { return val>a.val; } }; struct Edge { int to,next,cost; }; Edge E[MaxM]; int head[MaxN],Ecou; int vis[MaxN]; void Dijkstra(int lowcost[],int N,int start) { priority_queue <Node> que; Node temp; int u,v,c; for(int i=1;i<=N;++i) { lowcost[i]=INF; vis[i]=0; } que.push(Node(start,0)); lowcost[start]=0; while(!que.empty()) { temp=que.top(); que.pop(); u=temp.id; if(vis[u]) continue; vis[u]=1; for(int i=head[u];i!=-1;i=E[i].next) { v=E[i].to; c=E[i].cost; if(lowcost[v]>lowcost[u]+c && !vis[v]) { lowcost[v]=lowcost[u]+c; que.push(Node(v,lowcost[v])); } } } } void init(int N) { for(int i=1;i<=N;++i) head[i]=-1; Ecou=0; } void addEdge(int u,int v,int c) { E[Ecou].to=v; E[Ecou].cost=c; E[Ecou].next=head[u]; head[u]=Ecou++; } int ans[MaxN]; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int N,M; int a,b,c; scanf("%d %d",&N,&M); init(N); while(M--) { scanf("%d %d %d",&a,&b,&c); addEdge(a,b,c); } Dijkstra(ans,N,1); printf("%d\n",ans[N]); return 0; }
(简单) POJ 3159 Candies，Dijkstra+差分约束。
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原文地址：http://www.cnblogs.com/whywhy/p/4338505.html
