hdu 6201 transaction transaction transaction（最长路）

transaction transaction transaction

题目链接：transaction transaction transaction

题意：给你一棵树，每个点都有点权，每条边也有边权，要求选择起点S和终点T，使得val点T−val点S−val边最大

思路：建立源点和汇点，源点到树上所有点的权值为p[i]，树上所有点到汇点的权值为−p[i]，然后直接建图，spfa跑最长路即可

代码：

#include<bits/stdc++.h>using namespace std;const int maxn=1e5+10;struct edge{    int v,w,next;} E[maxn<<2];int inq[maxn],dis[maxn],first[maxn],p[maxn];int n,len;inline void scan_d(int &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')        ret = ret * 10 + (c - '0'), c = getchar();}int spfa(int st,int ed){    memset(inq,0,sizeof(inq));    memset(dis,0,sizeof(dis));    queue<int>q;    q.push(st),inq[st]=1;    while(!q.empty())    {        st=q.front();        q.pop(),inq[st]=0;        for(int i=first[st]; ~i; i=E[i].next)        {            int v=E[i].v,w=E[i].w;            if(dis[v]<dis[st]+w)            {                dis[v]=dis[st]+w;                if(!inq[v])                {                    inq[v]=1;                    q.push(v);                }            }        }    }    return dis[ed];}void add_edge(int u,int v,int w){    E[len].v=v,E[len].w=w,E[len].next=first[u],first[u]=len++;}int main(){    int t;    scan_d(t);    while(t--)    {        memset(first,-1,sizeof(first));        scan_d(n);        for(int i=1; i<=n; ++i)            scan_d(p[i]);        len=0;        for(int i=1; i<=n; ++i)            add_edge(0,i,p[i]),add_edge(i,n+1,-p[i]);        int u,v,w;        for(int i=1; i<n; ++i)        {            scan_d(u),scan_d(v),scan_d(w);            add_edge(u,v,-w),add_edge(v,u,-w);        }        printf("%d\n",spfa(0,n+1));    }    return 0;}

直接跑n边spfa竟然没超时

代码：

#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>using namespace std;const int maxn=1e5+10;struct edge{    int v,w,next;} E[maxn<<1];int p[maxn],d[maxn],first[maxn];int n,len;inline void scan_d(int &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')        ret = ret * 10 + (c - '0'), c = getchar();}void spfa(int u){    queue<int>q;    q.push(u);    while(!q.empty())    {        u=q.front();        q.pop();        for(int i=first[u]; ~i; i=E[i].next)        {            int v=E[i].v,w=E[i].w;            if(d[v]<d[u]+w)            {                d[v]=d[u]+w;                q.push(v);            }        }    }}void add_edge(int u,int v,int w){    E[len].v=v,E[len].w=w,E[len].next=first[u],first[u]=len++;}int main(){    int t;    scan_d(t);    while(t--)    {        scan_d(n);        for(int i=1; i<=n; ++i)            scan_d(p[i]);        memset(first,-1,sizeof(first));        len=0;        int u,v,w;        for(int i=1; i<n; ++i)        {            scan_d(u),scan_d(v),scan_d(w);            add_edge(u,v,p[v]-p[u]-w);            add_edge(v,u,p[u]-p[v]-w);        }        memset(d,0,sizeof(d));        for(int i=1; i<=n; ++i)            spfa(i);        int maxx=0;        for(int i=1; i<=n; ++i)            maxx=max(maxx,d[i]);        printf("%d\n",maxx);    }    return 0;}