题目链接：http://poj.org/problem?id=3159
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Candies
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 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 Input The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 throughN. Output Output one line with only the largest difference desired. The difference is guaranteed to be finite. Sample Input 2 2 1 2 5 2 1 4 Sample Output 5 Hint 32-bit signed integer type is capable of doing all arithmetic. Source POJ Monthly--2006.12.31, Sempr |
最短路径的经典题，用堆优化一下就完了。。算做是稍微复习下吧。
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<cstdlib>
#define CLR(A) memset(A,0,sizeof(A))
using namespace std;
struct Node{
int k,w;
};
bool operator <(const Node &a,const Node &b){return a.w>b.w;}
priority_queue<Node> q;
vector<vector<Node> > g;
bool used[30010]={0};
int main(){
int n,m;
Node p;
while(~scanf("%d%d",&n,&m)){
g.clear();
g.resize(n+1);
CLR(used);
for(int i=1;i<=m;i++){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
p.k=b;p.w=c;
g[a].push_back(p);
}
p.k=1;p.w=0;
q.push(p);
while(!q.empty()){
p=q.top();q.pop();
if(used[p.k]) continue;
used[p.k]=true;
if(p.k==n) break;
Node tmp;
for(int i=0;i<g[p.k].size();i++){
tmp.k=g[p.k][i].k;
if(used[tmp.k]) continue;
tmp.w=p.w+g[p.k][i].w;
q.push(tmp);
}
}
printf("%d\n",p.w);
}
return 0;
}