woj1009 最短路 The Legend of Valiant Emigration

---

title: woj1009 最短路 The Legend of Valiant Emigration

date: 2020-03-07

categories: acm

tags: [acm,最短路,woj]

SPFA，套了模板就很简单。输入处理一下。

description

The holy topic of human being,love,appears now.Firstly,I'd like to introduce a love song cited from <>

to you as follows.

The Woman

Your lips cover me with kisses;

your love is better than wine.

There is a fragrance about you;

the sound of your name recalls it.

No woman could help loving you.

Take me with you,and we'll run away.

be my king and take me to your room.

We will be happy together,

drink deep,and lose ourselves in love.

No wonder all women love you.

Women of Jerusalem,I am dark but beautiful,

dark as the desert tents of Kedar,

but beautiful as the curtains in Soloman's palace.

Don't look down on me because of my colour,

because the sun has tanned me.

My brotherw were angry with me

and made me work in the vineyard.

I had no time to care for myself.

Tell me,my love,

Where will you lead your flock to graze?

Where will they rest from the noonday sun?

Why should I need to look for you

among the flocks of the other shepherds?

The Man

Don't you know the place,loveliest of women?

Go and follow the flock;

find pasture for your goats near the tents of the shepherds.

You,my love,excite men as a mare excites the stallions of

Pharaoh's chariots.

Your hair is beautiful upon your cheeks

and falls along your neck like jewels.

But we will make for you a chain of gold with ornaments of silver.

The Woman

My king was lying on his couch,

and my perfume filled the air with fragrance.

My lover has the scent of myrrh as he lies upon my breasts.

My lover is like the wild flowers

that bloom in the vineyards at Engedi.

The Man

How beautiful you are,my love;

how your eyes shine with love!

The Woman

How handsome you are,my dearest;

how you delight me!

The green grass will be our bed;

the cedars will be the beams of our house,

and the cypress-trees the ceiling.

I am only a wild flower in Sharon,

a lily in a mountain valley.

The Man

Like a lily among thorns

is my darling among women.

The Woman

Like an apple-tree among the trees of the forest,

so is my dearest compared with other men.

I love to sit in its shadow,

and its fruit is sweet to my taste.

He brought me to his banqueting hall

and raised the banner of love over me.

Restore my strength with raisins

and refresh me with apples!

I am weak from passion.

His left hand is under my head,

and his right hand caresses me.

Promise me,women of Jerusalem;

swear by the swift deer and the gazelles

that you will not interrupt our love.

How beautiful the song is! The God is moved by the holy love between them and decides to help them run away to that place

having full-blown the green grass,the cedars,the cypress-trees and the flowers .There are many cities at that time,such as

Jerusalem,Bethlehem,Carmel,Hebron,Ziklog,Beersheba,Gath,Libnah,Ekron,Beth Horon,Shiloh,Gilgal,and so on.The two lovers

sets out from Jerusalem,and the destination is The Garden of Eden.There are roads between cities,each road has a

guard who has "pguard" power to beat and "sguard" speed to run.The two lovers can run away together,they have "plovers"

power to beat and "slovers" speed to run.  They can go through the road if and only if their power "plovers" is bigger than the guard and the speed "slovers"

is bigger than the guard..The God help them calculate the cost of each road,and place a

heart-shaped elixir,associated with a letter to distinguish them,in each road.Your task is to minimize the cost

and display the letters of elixirs they will get in the way.

输入格式

There will be several test cases.The first line contains two integers n(1<=n<=100) indicates the number of cities,and m(1<=m<=4000)

indicates the number of roads between cities.The next m lines list the power and speed of the guard,the cost and the letter of the elixir

the God placed for each road.Each line has the form:

u v pguard sguard cost letter while u and v represents the edge ,pguard and sguard represents power and speed of the guard, cost represents the cost of

the road and letter represents the letter of the elixir the God placed.The last line contains

plovers slovers. The city numberd 0 represents Jerusalem,while n-1 represents The Garden of Eden.

输出格式

For each test case,you should output the letters of the elixirs they get in the way which minimize the cost.

There'll be an empty line following each test case.

样例输入

5 5

0 1 10 10 10 L

0 2 20 14 10 A

1 2 10 10 20 o

2 3 10 10 10 V

3 4 5 30 10 E

14 35

样例输出

LoVE

code

//输入n m n cities m条边
//格式 u v pguard sguard cost letter
//while u and v represents the edge ,pguard and sguard represents power and speed of the guard,
// cost represents the cost of the road and letter represents the letter of the elixir the God placed.
//0 1 10 10 10 L
//The last line contains plovers slovers.plovers要大于路上守卫的guard,slovers要大于守卫的speed
// n 1-100 m 1-4000
//There'll be at least one path to go from Jerusalem to The Garden of Eden.You can assume no two paths have the same lentgh .

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<queue>
#include<cstring>

using namespace std;
#define INF 0x3f3f3f3f

char word[4005];
int plovers,slovers;
int n, m,edgenum;
const int maxn=105;  //const    点的数

struct Edge {
int from, to, power,speed,dist;
char word;
Edge(int u, int v,int p,int s, int d,char word):from(u),to(v),power(p),speed(s),dist(d),word(word){}
};

vector<Edge> edges;
vector<int> G[maxn];
bool inq[maxn]; //是否在队列里
int d[maxn]; //s到各个点的距离
int p[maxn]; //最短路中的上一条弧
int cnt[maxn]; //入队次数 次数大于n则说明有负环

void init(int n) {
edgenum=0;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}

void AddEdge(int from, int to, int power,int speed,int dist,char word) {
edges.push_back(Edge(from, to, power,speed,dist,word));
edgenum = edges.size();           //错误，一开始写成m，而m又是全局变量导致下面data()的循环除了问题
G[from].push_back(edgenum-1);  //边的标号
}

//拓扑排序可以判断有向图有没有环
//dfs判断图联通

bool bellman_ford(int s) {  ///BF 适用于含有负边的图。如果有负边，返回false。Dijkstra算法无法判断含负权边的图的最短路.二者都适用于有向有环图
    queue<int> Q;
    memset(inq, 0, sizeof(inq));
    memset(cnt, 0, sizeof(cnt));
    for(int i = 0; i < n; i++) d[i] = INF;
    d[s] = 0;
    inq[s] = true;
    p[s]=-1;     //加了一条
    Q.push(s);
    while(!Q.empty()) {
        int u = Q.front(); Q.pop();
        inq[u] = false;
        for(int i = 0; i < G[u].size(); i++) {
            Edge& e = edges[G[u][i]];
                if(d[u] < INF && d[e.to] > d[u] + e.dist&&plovers>e.power&&slovers>e.speed) {    //模板，+条件
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = G[u][i];
                    if(!inq[e.to]) { Q.push(e.to); inq[e.to] = true; if(++cnt[e.to] > n) return false;}
                }
        }
    }
    vector<char>ans;  //也可以写一个函数递归,没必要
    int tmp=p[n-1];
    while(tmp!=-1){
        ans.push_back(edges[tmp].word);
        tmp=p[edges[tmp].from];
    }
    for(int i=ans.size()-1;i>-1;i--)
        cout<<ans[i];
    cout<<endl;
    return true;
}

void data(){
    init(n);
    int from,to,power,speed,dist,tmp;
    char word;
    for(int i=0;i<m;i++){
        /*  这样写有问题。读了空格
           scanf("%d%d%d%d%d",&from,&to,&power,&speed,&dist);
           scanf("%c",&word);
           */
          /*
          scanf("%d%d%d%d%d",&from,&to,&power,&speed,&dist);
           getchar();scanf("%c",&word);getchar();*/     //处理最后一个字符前的空格这样可以，但是麻烦
           scanf("%d %d %d %d %d %c",&from,&to,&power,&speed,&dist,&word);  //注意%c不是%s 。也可以fstream
           AddEdge(from,to,power,speed,dist,word);
           }
    cin>>plovers>>slovers;
    bellman_ford(0);
}

int main(){

    while(scanf("%d%d",&n,&m)==2){
        data();
        getchar();  //输入有空行
    }
    return 0;
}