POJ3662 Telephone Lines( dijkstral + 二分 )

　　POJ3662 Telephone Lines

　　　　题目大意:要在顶点1到顶点n之间建一条路径,假设这条路径有m条边,其中有k条边是免费的,剩余m-k条边是要收费的,

　　　　　　　　求这m-k条边中花费最大的一条边的最小花费.

　　　　让m条边中原本花费最大的k条边成为免费的边,则这时m-k条边中花费最大的一条边的花费最小.

　　　　二分枚举m-k条边中花费最大的一条边的最小花费x,dijkstra求最短路径时,将花费大于x的边的花费设为1(花费为INF的边不变),花费小于等于x的边设为

0,则d[v-1]中返回的就是花费大于x的边数,当返回值小余等于k时,说明mid小了,ub=mid,否则,lb=mid+1;

　　最后输出mid或lb即可

　　　　一开始我的dijkstra未用队列优化,954ms飘过,用邻接矩阵存储时一开始一定要把所有边都初始化为INF,对cost[v][u]判断时,花费为INF的边不变

　　　　未优化的dijkstra

　　　　

/*
* Created:     2016年04月03日 14时11分34秒 星期日
* Author:      Akrusher
*
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define in3(x1,x2,x3) scanf("%d%d%d",&(x1),&(x2),&(x3))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

int cost[1005][1005];
int d[1005];
bool used[1005];
int V;
const int MAX_L=1000000;
int dijkstra(int s,int x){
    int pay;
    fill(d,d+V,INF);
    fill(used,used+V,false);
    d[s]=0;
    while(true){
        int v=-1;
        for(int u=0;u<V;u++){ //从未使用的顶点中选取一个距离最小的顶点
            if(!used[u]&&(v==-1||d[v]>d[u])) v=u; //注意此处是v=u;(wa了半天)
        }
        if(v==-1) break; //所有的顶点都被使用过了;
            used[v]=true;

            rep(u,0,V){    //更新最短距离
            if(cost[v][u]>x&&cost[v][u]!=INF) pay=1;//注意此处的判断,大于x的还有可能为INF,此处是pay=1,不是cost[v][u]=1;
            else if(cost[v][u]<=x) pay=0;  //等于x的电缆线不需要花钱,也为所求的最大花费的电缆线
            else if(cost[v][u]==INF) pay=INF; //一开始忘了判断这个,wa的不要不要的
            d[u]=min(d[u],d[v]+pay);
        }
    }
return d[V-1];//返回的是比x大的个数
}
int main()
{
    int n,p,k,a,b,c,lb,ub,mid;
    in3(n,p,k);
    V=n;
    fill(cost[0],cost[0]+1005*1005,INF);
    rep(i,0,p){
        in3(a,b,c);
        a--;b--;
        cost[a][b]=c;
        cost[b][a]=c;
    }
    lb=0,ub=MAX_L+1;//边的最大值为1000000
    while(ub>lb){  //二分时mid总是向下取整,区间小的时候让lb=mid+1,相等时让ub=mid则不会陷入死循环
        mid=(ub+lb)>>1;//位移运算符更高效
        if(dijkstra(0,mid)<=k) ub=mid;
        else lb=mid+1;
    }
    if(lb==MAX_L+1) printf("-1\n");
    else out(ub);
    return 0;
}

一下是用优先队列优化,时间是125ms;

　　

/*
* Created:     2016年04月03日 14时11分34秒 星期日
* Author:      Akrusher
*
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define in3(x1,x2,x3) scanf("%d%d%d",&(x1),&(x2),&(x3))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

struct edge{int to,cost;};
typedef pair<int,int> P; //first是最短距离,second是顶点编号
int V;
vector<edge> G[1005];
int d[1005];
const int MAX_L=1000000;
int dijkstra(int s,int x){
    priority_queue <P,vector<P>,greater<P> > que;
    fill(d,d+V,INF);
    d[s]=0;
    que.push(P(0,s));
    while(!que.empty()){
        P p=que.top();que.pop();
        int v=p.second;
        if(d[v]<p.first) continue;
        rep(i,0,G[v].size()){
            edge e=G[v][i];
            if(e.cost>x) e.cost=1;
            else e.cost=0;
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                que.push(P(d[e.to],e.to));
                }
            }
        }
return d[V-1];//返回的是比x大的个数
}
int main()
{
    int n,p,k,a,b,c,lb,ub,mid;
    in3(n,p,k);
    V=n;

    rep(i,0,p){
        in3(a,b,c);
        a--;b--;
        edge e;
        e.to=b;
        e.cost=c;
        G[a].push_back(e);
        e.to=a;
        e.cost=c;
        G[b].push_back(e);//没有重边才能这样赋值
    }
    lb=0,ub=MAX_L+1;//边的最大值为1000000
    while(ub>lb){  //二分时mid总是向下取整,区间小的时候让lb=mid+1,相等时让ub=mid则不会陷入死循环
        mid=(ub+lb)>>1;//位移运算符更高效
        if(dijkstra(0,mid)<=k) ub=mid;
        else lb=mid+1;
    }
    if(lb==MAX_L+1) printf("-1\n");
    else out(ub);
    return 0;
}

　　　　

时间： 2024-10-28 23:00:09

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