sicily 1003 Go To School

题目链接 http://soj.me/show_problem.php?pid=1003&cid=892

Description

Lucy lives in city GZ. There are N (2 <= N && N <= 100) towns in GZ。There are M (1 <= M <= 1000) two-way roads connecting the towns, and every road has a length w (1 <= w <= 1000). Lucy’s home is in town 1, and the school is in town N. Every day, Lucy goes to school on time, she always takes the shortest path. Please write a program to calculate the length of the shortest path from town 1 to town N.

Input

The input contains several test cases.

The first line contains two integers N, M. The following M lines each contains three integers x, y, w, meaning there is a two-way road between town x and y, and the length of the road is w.

The input will be terminated by EOF.

Output

Output the length of the shortest path from town 1 to town N. It is guaranteed that there is at least one path from town 1 to town N.

Sample Input

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2 12 1 4793 31 3 101 2 42 3 4

Sample Output

4798

这题比较传统，找最短路径，所以知道求最短路径算法的话，这题还是不难的。我用的是dijkstra算法。

#include <iostream>#include <cstring>using namespace std;int edge[110][110];int const my_MAX = 9999;int dijkstra(int s, int d){    int dis[110];    bool vis[110];    int v, i;    memset(dis, my_MAX, sizeof(dis));    dis[0] = 0;    dis[s] = 0;    memset(vis, false, sizeof(vis));        while (true) {        int min = my_MAX;        for (i = 1; i <= d; i++) {            if (!vis[i] && dis[i] < min) {                v = i;                min = dis[i];            }        }                       if (min == my_MAX)            break;                  vis[v] = true;          dis[v] = min;        for (i = 1; i <= d; i++) {            if (!vis[i] && dis[i] > edge[v][i] + dis[v])                dis[i] = edge[v][i] + dis[v];        }    }    return dis[d];}int main(){   //  freopen("input.txt","r",stdin);    int n, m;    while (cin >> n >> m) {        int x, y, d;        memset(edge, my_MAX, sizeof(edge));        for (int i = 0; i < m; i++) {            cin >> x >> y >> d;                     edge[x][y] = d;            edge[y][x] = d;        }        cout << dijkstra(1,n) << endl;    }    return 0;}