upc组队赛6 Bumped!【最短路】
Bumped!
题目描述
Peter returned from the recently held ACM ICPC World finals only to find that his return flight was overbooked and he was bumped from the flight! Well, at least he wasn’t beat up by the
airline and he’s received a voucher for one free flight between any two destinations he wishes.
He is already planning next year’s trip. He plans to travel by car where necessary, but he may be using his free flight ticket for one leg of the trip. He asked for your help in his planning.
He can provide you a network of cities connected by roads, the amount it costs to buy gas for traveling between pairs of cities, and a list of available flights between some of those cities. Help Peter by finding the minimum amount of money he needs to spend to get from his hometown to next year’s destination!
输入
The input consists of a single test case. The first line lists five space-separated integers n, m, f, s, and t, denoting the number of cities n (0 < n ≤ 50 000), the number of roads m (0 ≤ m ≤ 150 000), the number of flights f (0 ≤ f ≤ 1 000), the number s (0 ≤ s < n) of the city in which Peter’s trip starts, and the number t (0 ≤ t < n) of the city Peter is trying to travel to. (Cities are numbered from 0 to n − 1.)
The first line is followed by m lines, each describing one road. A road description contains three space-separated integers i, j, and c (0 ≤ i, j < n, i 6= j and 0 < c ≤ 50 000), indicating there is a road connecting cities i and j that costs c cents to travel. Roads can be used in either direction for the same cost. All road descriptions are unique.
Each of the following f lines contains a description of an available flight, which consists of two space-separated integers u and v (0 ≤ u, v < n, u 6= v) denoting that a flight from city u to city v is available (though not from v to u unless listed elsewhere). All flight descriptions are unique.
输出
Output the minimum number of cents Peter needs to spend to get from his home town to the competition,using at most one flight. You may assume that there is a route on which Peter can reach his destination.
样例输入
8 11 1 0 5
0 1 10
0 2 10
1 2 10
2 6 40
6 7 10
5 6 10
3 5 15
3 6 40
3 4 20
1 4 20
1 3 20
4 7
样例输出
45
题意
n个点,m条边,f条飞行路线,s起始点,t终点
其中f条飞行路线可以使一条边的距离为0;
题解
这题很坑,相当坑,有如下几点
不管f有多少条 ,只能选一条 ,所以我们枚举每一条f边为0,去求最短路;
如果f=0,别直接不跑了
数据范围是long long ,如果初始化为INF 就会WA!! 太坑了! 改成9999999999或更多(在此WA了几百发,留下惨痛教训)
可以用map或者set来存
代码
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,a,n) for(int i=a;i<n;++i)
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define sca(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<ll,ll> P;
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 105;
const int maxn =1000010;
using namespace std;
struct node{
ll to;
ll cost;
bool operator < (node b)const{
return cost > b.cost;
}
};
struct edge{
ll to;
ll cost;
};
vector<edge> E[maxn];
int vis[maxn];
ll dis[maxn];
priority_queue<node>q;
ll x,y,z;
ll n,m,f,st,ed;
void Dijkstra(ll s){
memset(vis,0,sizeof vis);
for(int i = 0; i < n; i++)
dis[i] = 9999999999;
while(!q.empty())
q.pop();
dis[s] = 0;
q.push(node{s,0});
node tmp;
while(!q.empty()){
tmp = q.top();
q.pop();
ll u = tmp.to;
if(vis[u])
continue;
vis[u] = 1;
for(int i = 0; i < E[u].size(); i++){
ll v = E[u][i].to;
ll w = E[u][i].cost;
if(!vis[v] && dis[v] > dis[u] + w){
dis[v] = dis[u] + w;
q.push(node{v,dis[v]});
}
}
}
}
int main(){
scanf("%lld%lld%lld%lld%lld",&n,&m,&f,&st,&ed);
while(m--){
scanf("%lld%lld%lld",&x,&y,&z);
E[x].pb(edge{y,z});
E[y].pb(edge{x,z});
}
ll ans = 9999999999;
Dijkstra(st);
ans = dis[ed];
while(f--){
scanf("%lld%lld",&x,&y);
E[x].pb(edge{y,0});
Dijkstra(st);
ans = min(ans,dis[ed]) ;
E[x].erase(E[x].end() - 1);
}
printf("%lld\n",ans);
return 0;
}
/*
6 7 2 0 5
0 1 10
0 3 5
1 4 10
2 3 5
0 5 100
4 5 5
2 5 5
0 2
1 3
*/
upc组队赛6 Bumped!【最短路】的更多相关文章
- upc组队赛15 Made In Heaven【第K短路 A*】
Made In Heaven 题目描述 One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with ...
- upc组队赛3 Chaarshanbegaan at Cafebazaar
Chaarshanbegaan at Cafebazaar 题目链接 http://icpc.upc.edu.cn/problem.php?cid=1618&pid=1 题目描述 Chaars ...
- upc组队赛1 过分的谜题【找规律】
过分的谜题 题目描述 2060年是云南中医学院的百年校庆,于是学生会的同学们搞了一个连续猜谜活动:共有10个谜题,现在告诉所有人第一个谜题,每个谜题的答案就是下一个谜题的线索....成功破解最后一个谜 ...
- upc组队赛1 不存在的泳池【GCD】
不存在的泳池 题目描述 小w是云南中医学院的同学,有一天他看到了学校的百度百科介绍: 截止到2014年5月,云南中医学院图书馆纸本藏书74.8457万册,纸质期刊388种,馆藏线装古籍图书1.8万册, ...
- upc组队赛1 黑暗意志【stl-map】
黑暗意志 题目描述 在数千年前潘达利亚从卡利姆多分离之时,迷雾笼罩着这块新形成的大陆,使它不被外来者发现.迷雾同样遮蔽着这片大陆古老邪恶的要塞--雷神的雷电王座.在雷神统治时期,他的要塞就是雷电之王力 ...
- upc组队赛1 闪闪发光 【优先队列】
闪闪发光 题目描述 一所位于云南昆明的中医药本科院校--云南中医学院. 因为报考某专业的人数骤减,正面临着停招的危机. 其中有九名少女想到一条妙计--成为偶像, 只要她们成为偶像,学校的名气便会增加, ...
- upc组队赛1 流连人间的苏苏
流连人间的苏苏 题目描述 苏苏在做红尘仙的任务时,发现坐落于风景秀丽.四季如春的昆明市的云南中医学院. 没过多久,苏苏就喜欢上了这个学校.以致于苏苏忘了回涂山的时间,现在她只剩下d天的时间待在云南中医 ...
- upc组队赛18 THE WORLD【时间模拟】
THE WORLD 题目链接 题目描述 The World can indicate world travel, particularly on a large scale. You mau be l ...
- upc 组队赛18 STRENGTH【贪心模拟】
STRENGTH 题目链接 题目描述 Strength gives you the confidence within yourself to overcome any fears, challeng ...
随机推荐
- Java学习之多态---类成员变化
类成员 一.成员变量 编译时:变量(f)所属类(Fu)中是否有成员变量,有:编译成功,没有:编译失败 运行时:变量(f)所属类(Fu)中是否有成员变量,运行该类(Fu)中的成员变量 class Fu ...
- python Condition类(锁)
Condition(条件变量)通常与一个锁关联.需要在多个Contidion中共享一个锁时,可以传递一个Lock/RLock实例给构造方法,否则它将自己生成一个RLock实例. 不理解锁的,请看上一条 ...
- Python for Eclipse插件Pydev安装后eclipse里面看不到
反复安装几次在eclipse里面都找不到,崩溃了,有的说是pydev版本不对,应该和python版本一致,都是扯淡,去pydev官网看了下requirement原来是需要Java1.7+以上,安装1. ...
- js 可迭代对象
作用:可以简化使用循环语句初始化一个变量记录迭代位置的操作 function createIterator(iterms) { let i = 0 return { next() { let done ...
- XMPP即时通讯协议使用(十一)——Openfire表结构汇总
行号 字段名称 字段描述 字段类型 长度 主键 说明 允许为空 用户组数据表(ofGroup) 1 groupName 组名 varchar2 50 ★ NOT NULL 2 descriptio ...
- 使用Maven创建Web项目(转)
转自:http://ju.outofmemory.cn/entry/49508 本文通过Maven完成一个简单的Web项目(注意,Spring配置不是重点,看看就行) 1.从Maven模板创建Web应 ...
- Linux下tree的使用介绍
1.在ubuntu系统中默认是没有tree这个命令的,需要安装,用下面的命令就可以安装tree这个命令工具sudo apt-get install tree. 2.首先来说说使用tree这个命令,就是 ...
- css3--css3模块
CSS3被拆分为"模块".旧规范已拆分成小块,还增加了新的. 一些最重要CSS3模块如下: 1.选择器(.基本选择器,层次选择器,伪类选择器,伪元素,属性选择器,通配符) 2.盒模 ...
- MS Word2016加载EndnoteX6插件
我的软件环境是:Win10 x64.MS Office 2016 x64.Endnote X6 32位. 在安装完MSO和Endnote后,Word中未能自动加载Endnote插件.现将启用方法记录如 ...
- 一次峰回路转的getshell
扫目录发现 http://www.xxx.test.cn/bak/以及/bak/upload.jsp