Fishmonger SPOJ - FISHER 有限制的最短路

A fishmonger wants to bring his goods from the port to the market. On his route he has to traverse an area with many tiny city states. Of course he has to pay a toll at each border.

Because he is a good business man, he wants to choose the route in such a way that he has to pay as little money for tolls as possible. On the other hand, he has to be at the market within a certain time, otherwise his fish start to smell.

Input

The first line contains the number of states n and available time t. The first state is the port, the last state is the market. After this line there are n lines with n numbers each, specifying for each state the travel time to the i-th state. This table is terminated with an empty line. The table of the tolls follows in the same format.

n is at least 3 and at most 50. The time available is less than 1000. All numbers are integers.

There are many test cases separated by an empty line. Input terminates with number of states and time equal 0 0.

Output

For each test case your program should print on one line the total amount of tolls followed by the actual travelling time.

Example

Sample input:
4 7
0 5 2 3
5 0 2 3
3 1 0 2
3 3 2 0

0 2 2 7
2 0 1 2
2 2 0 5
7 2 5 0

0 0

Sample output:
6 6
      

This corresponds to the following situation, the connections are labeled with (time, toll):

圖中邊為有向邊， 每條邊的資訊有 起點到終點所消耗的時間、金錢。

輸出 n, k    節點個數與限制時間。

求在限制時間内最少消耗多少金錢從起點(1)到終點（n）

n~[3,50], time<=1000,

當n k 同時為0時， 輸入結束。

1. 節點資料量小 可以考慮DFS

#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ms(x) memset(x, 0, sizeof(x))
using namespace std;
const int N = ;
int tim[N][N], cost[N][N];
bool vis[N];
int n, k;
int nt, nc;
void dfs(int x, int tpcost, int tptime){
    if(x == n){
        if(tpcost < nc ){
            nc = tpcost;
            nt = tptime;
        }
        return;
    }
    for(int i=;i<n;i++){
        if(vis[i]) continue;
        if(tpcost + cost[x][i]<=nc && tptime + tim[x][i] <= k){
            vis[i] = ;
            dfs(i, tpcost + cost[x][i],tptime + tim[x][i]);
            vis[i] = ;
        }
    }
}
int main()
{
    while(scanf("%d%d", &n, &k)!=EOF){
        if(!n && !k) break;
        nt = nc = ;
        ms(vis);
        for(int i=;i<n;i++){
            for(int j=;j<n;j++){
                scanf("%d",&tim[i][j]);
            }
        }
        for(int i=;i<n;i++){
            for(int j=;j<n;j++){
                scanf("%d",&cost[i][j]);
            }
        }
        if(tim[][n]<=k){
            nt = tim[][n];
            nc = cost[][n];
        }
        else {nt = k, nc = inf;}
        vis[] = ;
        dfs(,,);
        printf("%d %d\n", nc, nt);
    }
    return ;
}
           

2. 最短路變形

将SPFA中的 dist[] vis[] 數組修改成二維的  二維存時間。

跑SPFA時所用隊列 存pair

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ll long long
#define ms(x) memset(x, 0, sizeof(x))
using namespace std;
const int N = 60;
struct Edge{
    int v, cost;
    Edge(int _v = 0, int _cost = 0):v(_v),cost(_cost){}
};
vector<Edge>E[N];
void addedge(int u, int v, int w){
    E[u].push_back(Edge(v, w));
}
bool vis[N][2005];
int dist[N][2005];
int tim[N][N];
int m;
void SPFA(int st, int n){
    ms(vis);
    memset(dist, inf, sizeof(dist));
    queue<pair<int,int> >que;
    que.push(make_pair(st, 0));
    dist[st][0] = 0;
    while(!que.empty()){
        int u = que.front().first;
        int t = que.front().second;
        vis[u][t] = false;
        que.pop();
        for(int i=0;i<E[u].size();i++){
            int v = E[u][i].v;
            int tptime = t + tim[u][v];
            if(tptime>1000) continue;
            if(dist[v][tptime] > dist[u][t] + E[u][i].cost){
                dist[v][tptime] = dist[u][t] + E[u][i].cost;
                if(!vis[v][tptime]){
                    vis[v][tptime] = true;
                    que.push(make_pair(v, tptime));
                }
            }
        }
    }
    int ans = inf, tim;
    for(int i=0;i<=m;i++){
        if(dist[n-1][i] < ans){
            ans = dist[n-1][i];
            tim = i;
        }
        //printf("%d %d\n",dist[n-1][i], i);
    }
    printf("%d %d\n",ans, tim);
}
int main()
{
    int n;
    while(scanf("%d%d", &n,&m)!=EOF){
        for(int i=0;i<N;i++){
            if(!E[i].empty()) E[i].clear();
        }
        if(!n && !m) break;
        int tpx;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d", &tim[i][j]);
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d", &tpx);
                if(i==j) continue;
                addedge(i, j, tpx);
            }
        }
        SPFA(0, n);
    }
    return 0;
}