HDU4571/2013年长沙赛区Travel in time

Description

Bob gets tired of playing games, leavesAlice, and travels to Changsha alone.YueluMountain, Orange Island, Window of the World, theProvincial Museum etc...are scenic spots Bob wants to visit. However, his timeis very limited, he can’t visit them all.

Assuming that there areN scenicspots in Changsha, Bob defines a satisfaction valueSi to each spot. Ifhe visits this spot, his total satisfaction value will plusSi. Bobhopes that within the limited timeT, he can start at spotS, visitsome spots selectively, and finally stop at spotE, so that the totalsatisfaction value can be as large as possible. It's obvious that visiting thespot will also cost some time, suppose that it takesCi units of time tovisit spoti ( 0 <= i < N ).

Always remember, Bob can choose to pass bya spot without visiting it (includingS andE), maybe he justwant to walk shorter distance for saving time.

Bob also has a special need which is thathe will only visit the spot whose satisfaction value isstrictly largerthan that of which he visited last time. For example, if he has visited a spotwhose satisfaction value is 50, he would only visit spot whose satisfactionvalue is 51 or more then. The paths between the spots are bi-directional, ofcourse.

 

Input

The first line is an integerW,which is the number of testing cases, and the W sets of data are following.

The first line of each test data containsfive integers:N M T S E.N represents the number of spots, 1< N < 100;M represents the number of paths, 0 <M< 1000; T represents the timelimitation, 0 <T <= 300;S means the spot Bob starts from.E indicates the end spot. (0 <=S,E < N)

The second line of the test data containsNintegersCi ( 0 <= Ci <= T ), which means the costof time if Bob visits the spoti.

The third line also hasN integers,which means the satisfaction valueSi that can be obtained by visitingthe spoti ( 0 <= Si < 100 ).

The next M lines, each line containsthree integersu v L, means there is a bi-directional pathbetween spotu and v and it takes L units of time to walkfrom u tov or fromv to u. (0 <= u, v< N, 0 <=L <=T)

 

 

 

Output

Output case number in the first line (formattedas the sample output).

The second line contains an integer, whichis the greatest satisfaction value.

If Bob can’t reach spotE inT units oftime, you should output just a “0” (without quotation marks).

 

Sample Input

1

4 4 22 0 3

1 1 1 1

5 7 9 12

0 1 10

1 3 10

0 2 10

2 3 10

 

 

Sample Output

Case #1:

21

先用floyd预处理出两两之间的距离，建立有向图，边的方向从满意度低的点到满意度高的点。注意构造两个抽象的起点和终点，表示不访问该点而经过该点。

再跑一遍二维spfa,dist[i][j]表示在j时刻到达i点得到的最大满意度

最后在dist[终点][0~T]以及dist[抽象终点][0~T]中找最大值

 

#include <iostream>#include <cstdio>#include <vector>#include <queue>using namespace std;#define MAXNODE 110#define MAXTIME 721#define MAXSATIS 100#define oo (1<<29)struct PATH{    int v, L;    PATH(int _v, int _L){v = _v; L = _L;}};// Data of the graphint N, M;vector<struct PATH> Graph[MAXNODE];// Start point, end point, time limitint Sp, Ep, T;// Satisfaction and time-cost of every sight spotint s[MAXNODE], c[MAXNODE];// Satisfaction[i][j] means the max satisfaction on Node i when time is jint Satisfaction[MAXNODE][MAXTIME];struct Node{    int position, time;    Node(){}    Node(int _position, int _time)    {        position = _position;        time     = _time;    }};queue<struct Node> que;// whether the node is in the queuebool InQueue[MAXNODE][MAXTIME];Node Pop_Queue(){    struct Node now = que.front();    que.pop();    InQueue[now.position][now.time] = false;    return now;}void Push_Queue(Node NewNode){    if (InQueue[NewNode.position][NewNode.time]) return;    que.push(NewNode);    InQueue[NewNode.position][NewNode.time] = true;}void SPFA(){    // initialize    while (!que.empty()) que.pop();    for (int i = 0; i < N + 2; ++i)        for (int j = 0; j <= T; ++j)        {            Satisfaction[i][j] = 0;            InQueue[i][j] = false;        }    Push_Queue( Node(N, 0) );    Satisfaction[N][0] = 0;    while (!que.empty())    {        struct Node now = Pop_Queue();        // go to another spot - Relaxation operating        int size = Graph[now.position].size(), NewTime;        for (int i = 0; i < size; ++i)        {            int v = Graph[now.position][i].v;            NewTime = now.time + Graph[now.position][i].L;            if (NewTime > T) continue;            if (Satisfaction[now.position][now.time] + s[v] > Satisfaction[v][NewTime])            {                Satisfaction[v][NewTime] = Satisfaction[now.position][now.time] + s[v];                Push_Queue( Node(v, NewTime) );            }        }    }}int Matrix[MAXNODE][MAXNODE];// Read datavoid Init(){    scanf("%d %d %d %d %d", &N, &M, &T, &Sp, &Ep);    for (int i = 0; i < N; ++i) scanf("%d", &c[i]);    for (int i = 0; i < N; ++i) scanf("%d", &s[i]);    for (int i = 0; i < N; ++i)        for (int j = 0; j < N; ++j)            Matrix[i][j] = (i == j)? (0) : (oo);    while (M--)    {        int u, v, l;        scanf("%d %d %d", &u, &v, &l);        Matrix[u][v] = Matrix[v][u] = min(Matrix[u][v], l);    }}void Floyd_Warshall(){    for (int k = 0; k < N; ++k)        for (int i = 0; i < N; ++i)            for (int j = 0; j < N; ++j)                Matrix[i][j] = min(Matrix[i][j], Matrix[i][k] + Matrix[k][j]);}void Add_Edge(int from, int to, int cost){    Graph[from].push_back(PATH(to, cost));}void Build_Graph(){    s[N + 1] = 0;    for (int i = 0; i < N + 2; ++i) Graph[i].clear();    for (int i = 0; i < N; ++i)        for (int j = i + 1; j < N; ++j)            if (Matrix[i][j] < oo)            {                if (s[i] > s[j]) Add_Edge(j, i, Matrix[i][j] + c[i]);                else if (s[i] < s[j]) Add_Edge(i, j, Matrix[i][j] + c[j]);            }    //Super source - N    for (int i = 0; i < N; ++i)        if (i != Sp && Matrix[Sp][i] < oo)            Add_Edge(N, i, Matrix[Sp][i] + c[i]);    Add_Edge(N, Sp, c[Sp]);    //Super dest - N + 1    for (int i = 0; i < N; ++i)        if (i != Ep && Matrix[i][Ep] < oo)            Add_Edge(i, N + 1, Matrix[i][Ep]);}int main(){    int Case;    scanf("%d", &Case);    for (int i = 1; i <= Case; ++i)    {        Init();        Floyd_Warshall();        Build_Graph();        SPFA();        int maxS = 0;        for (int t = 0; t <= T; ++t)        {            maxS = max (maxS, Satisfaction[N + 1][t]);            maxS = max (maxS, Satisfaction[Ep][t]);        }        printf("Case #%d:\n%d\n", i, maxS);    }    return 0;}