poj --- 1724 最短路变形 Dij + 优先队列

题目链接

题意:要从1到N的最短路并且每一条路上有对应的花费要求,总花费不能大于给定的一个值K,输出最短路径的长度.
方法: 不仅要最短,还有花费限制,所以就想到用优先队列,来维护一个优先级关系,使得答案最优.

AC代码:

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<functional>
#include<vector>
#include<stack>
#include<map>
#include<cstdlib>
#define CLR(x) memset(x,0,sizeof(x))
#define ll long long int
#define PI acos(-1.0)
#define db double
#define mod 1000000007
using namespace std;
const int maxn = 1e4+5;
const int inf=1e9;
int k,n,r;
int cnt;
struct node
{
    int to,w,c,next;
}s[maxn];   //模拟链表.

struct cmp    //优先队列需要用.
{
    int id,dis,cost;
    bool operator < (const cmp  &a) const {     //以取&号的为准!
        if(a.dis == dis ) return a.cost < cost ;  //其次是花费小的优先.
        return a.dis < dis ;   //先让路最短的优先.
    }
};

int head[maxn];

void add(int u,int v, int w,int c)    //模拟过程.
{
    s[cnt].to=v,s[cnt].w=w,s[cnt].c=c;
    s[cnt].next=head[u];
    head[u]=cnt++;

}

void dij()    //dij + 优先队列优化!!! , 学会啊 !!!
{
    priority_queue<cmp>q;
    cmp a,b;
    int res=inf;
    a.id=1;      //初始化情况
    a.dis=0;
    a.cost=0;
    q.push(a);
    while(!q.empty()){
        b=q.top();
        q.pop();

        if(b.id == n){
            res=b.dis;     //已找到目标点,所以推出bfs.
            break;
        }

        for(int i=head[b.id];i!=-1;i=s[i].next){
            int to=s[i].to;
            int w=s[i].w;
            if(s[i].c + b.cost <= k){    //花费小于k.
                a.id = to;
                a.dis = b.dis + w;
                a.cost = b.cost + s[i].c ;
                q.push(a);
            }
        }
    }
    if(res == inf ) printf("-1\n");
    else printf("%d\n",res);
}
int main()
{
    scanf("%d %d %d",&k,&n,&r);
    cnt=0;
    memset(head,-1,sizeof(head));
    for(int i=0;i<r;i++){
        int u,v,w,c;
        scanf("%d %d %d %d",&u,&v,&w,&c);
        add(u,v,w,c);     //单向边,所以只需要加一次.
    }
    dij();
}