URAL 1004. Sightseeing Trip

Author: 2015-10-31

ACM

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1. 题目

http://acm.timus.ru/problem.aspx?space=1&num=1004

1004. Sightseeing Trip

Time limit: 0.5 second
Memory limit: 64 MB
There is a travel agency in Adelton town on Zanzibar island. It has decided to offer its clients, besides many other attractions, sightseeing the town. To earn as much as possible from this attraction, the agency has accepted a shrewd decision: it is necessary to find the shortest route which begins and ends at the same place.
Your task is to write a program which finds such a route. In the town there are N crossing points numbered from 1 to N andM two-way roads numbered from 1 to M. Two crossing points can be connected by multiple roads, but no road connects a crossing point with itself. Each sightseeing route is a sequence of road numbers y1, …, yk, k > 2. The road yi (1 ≤ i ≤ k − 1)connects crossing points xi and xi+1, the road yk connects crossing points xk and x1. All the numbers x1, …, xk should be different. The length of the sightseeing route is the sum of the lengths of all roads on the sightseeing route, i.e. L(y1) + L(y2) + … + L(yk) where L(yi) is the length of the road yi (1 ≤ i ≤ k). Your program has to find such a sightseeing route, the length of which is minimal, or to specify that it is not possible, because there is no sightseeing route in the town.

Input

Input contains T tests (1 ≤ T ≤ 5). The first line of each test contains two integers: the number of crossing points N and the number of roads M (3 ≤ N ≤ 100; 3 ≤ M ≤ N · (N − 1)). Each of the next M lines describes one road. It contains 3 integers: the number of its first crossing point a, the number of the second one b, and the length of the road l (1 ≤ a, b ≤ N; a ≠ b;1 ≤ l ≤ 300). Input is ended with a “−1” line.

Output

Each line of output is an answer. It contains either a string “No solution.” in case there isn’t any sightseeing route, or it contains the numbers of all crossing points on the shortest sightseeing route in the order how to pass them (i.e. the numbersx1 to xk from our definition of a sightseeing route), separated by single spaces. If there are multiple sightseeing routes of the minimal length, you can output any one of them.

Sample

input output 
5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
4 3
1 2 10
1 3 20
1 4 30
-1

1 3 5 2
No solution.
Problem Source: Central European Olympiad in Informatics 1999

2. 思路

M条路有N个交点，给出每条路的起止点和长度，求起止位置在同一点的最小环线。

使用Floyd求最小环。

3. 代码

#include <cstdio>

const int MAX_N = 101;
const int MAX_DIST = 30000;
typedef int Graph[MAX_N][MAX_N];

void solveP8e_SightseeingTrip();
void inputMap(Graph map, int n, int m);
int floyd(Graph map, int n, int result[]);

int main() {
    // freopen("test.txt", "r", stdin);
    solveP8e_SightseeingTrip();
    return 0;
}

Graph map;
int result[MAX_N];
void solveP8e_SightseeingTrip() {
    while (1) {
        int n, m;
        if (scanf("%d %d", &n, &m) == 1)
            return;

        inputMap(map, n, m);

        int cnt = floyd(map, n, result);

        if (cnt) {
            for (int i = 0; i < cnt; ++i) {
                printf("%d ", result[i]);
            }
            printf("\n");
        } else {
            printf("No solution.\n");
        }
    }
}

Graph dist, path;
int floyd(Graph map, int n, int result[]) {
    int currDist = MAX_DIST, cnt = 0;

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            dist[i][j] = map[i][j];
            path[i][j] = i;
        }
    }

    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i < k; ++i) {
            for (int j = 1; j < i; ++j) {
                if (dist[i][j] + map[i][k] + map[k][j] < currDist) {
                    currDist = dist[i][j] + map[i][k] + map[k][j];
                    int currPos = j;
                    cnt = 0;
                    while (currPos != i) {
                        result[cnt++] = currPos;
                        currPos = path[i][currPos];
                    }
                    result[cnt++] = i;
                    result[cnt++] = k;
                }
            }
        }

        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (dist[i][j] > dist[i][k] + dist[k][j]) {
                    dist[i][j] = dist[i][k] + dist[k][j];
                    path[i][j] = path[k][j];
                }
            }
        }
    }

    return cnt;
}

void inputMap(Graph map, int n, int m) {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            map[i][j] = MAX_DIST;
        }
    }

    for (int i = 0; i < m; ++i) {
        int a, b, l;
        scanf("%d %d %d", &a, &b, &l);
        map[a][b] = map[a][b] < l ? map[a][b] : l;
        map[b][a] = map[a][b];
    }
}