Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9052    Accepted Submission(s): 2383

Problem Description

These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: 
The cost of the transportation on the path between these cities, and

a certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities.
You must write a program to find the route which has the minimum cost.

Input

First is N, number of cities. N = 0 indicates the end of input.
The data of path cost, city tax, source and destination cities are given in the input, which is of the form:

a11 a12 ... a1N
a21 a22 ... a2N
...............
aN1 aN2 ... aNN
b1 b2 ... bN

c d
e f
...
g h

where aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form:

Output

From c to d :
Path: c-->c1-->......-->ck-->d
Total cost : ......
......

From e to f :
Path: e-->e1-->..........-->ek-->f
Total cost : ......

Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case.

Sample Input

5

0 3 22 -1 4

3 0 5 -1 -1

22 5 0 9 20

-1 -1 9 0 4

4 -1 20 4 0

5 17 8 3 1

1 3

3 5

2 4

-1 -1

0

Sample Output

From 1 to 3 :

Path: 1-->5-->4-->3

Total cost : 21

From 3 to 5 :

Path: 3-->4-->5

Total cost : 16

From 2 to 4 :

Path: 2-->1-->5-->4

Total cost : 17

Source

Asia 1996, Shanghai (Mainland China)

题意，给一张n*n的地图，-1表示不联通，正整数为边权，每个点有点权，给定起点s和终点t，s到t的代价是经过的边权和加上路径上除了s和t的点权和，求s到t的最短路并输出路径

思路：floyd，path[i][j]表示i到j路径上的第一个点

/*
ID: LinKArftc
PROG: 1385.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 100;

int mp[maxn][maxn];
int path[maxn][maxn];//path[i][j]记录从i到j的下一个点
int tax[maxn];
int n, s, t;

void floyd() {
    for (int k = 1; k <= n; k ++) {
        for (int i = 1; i <= n; i ++) {
            for (int j = 1; j <= n; j ++) {
                if (i == k || j == k) continue;
                int tmp = mp[i][k] + mp[k][j] + tax[k];
                if (mp[i][j] > tmp) {
                    mp[i][j] = tmp;
                    path[i][j] = path[i][k];
                } else if (mp[i][j] == tmp) {
                    path[i][j] = min(path[i][j], path[i][k]);
                }
            }
        }
    }
}

void print_path() {
    printf("Path: %d", s);
    int cur = s;
    while (cur != t) {
        cur = path[cur][t];
        printf("-->%d", cur);
    }
    printf("\n");
}

int main() {
    while (~scanf("%d", &n) && n) {
        for (int i = 1; i <= n; i ++) {
            for (int j = 1; j <= n; j ++) {
                scanf("%d", &mp[i][j]);
                if (mp[i][j] == -1) mp[i][j] = inf;//最好用inf替换，用-1的话特判很容易错
                else path[i][j] = j;
            }
        }
        for (int i = 1; i <= n; i ++) scanf("%d", &tax[i]);
        floyd();
        while (~scanf("%d %d", &s, &t)) {
            if (s == -1 && t == -1) break;
            printf("From %d to %d :\n", s, t);
            print_path();
            printf("Total cost : %d\n\n", mp[s][t]);
        }
    }

    return 0;
}