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#include <stdio.h>#include <queue>#include <iostream>using namespace std;#define INF 0xfffff //因为为了辨别是否有负权,所以INF不能开太大#define MAX 1100int dist[MAX], pre[MAX], path[MAX][MAX];bool sign[MAX];void initialize(int n) //初始化{for(int i=1; i<=n; i++){{//pre[i] = 0;dist[i] = INF; //将距离开始全变为最大sign[i] = false;}for(int j=1; j<=n; j++)path[i][j] = INF; //图初始}}void spfa(int n, int start) //无法计算负权,对边较多的更快速{/* for (int i=1; i<=n; ++i)//初始化{dist[i] = INF;sign[i] = false;}*/queue<int> Q;dist[start] = 0;sign[start] = true;Q.push(start);while (!Q.empty()){int temp = Q.front();Q.pop();for (int i=0; i<=n; ++i){if (dist[temp] + path[temp][i] < dist[i])//存在负权的话,就需要创建一个COUNT数组,当某点的入队次数超过V(顶点数)返回。{dist[i] = dist[temp] + path[temp][i];if (!sign[i]){Q.push(i);sign[i] = true;}//这个内循环可以判断所要权值相对应条件的值如dist[start];}}sign[temp] = false;}}void floyd (int n) //边较为少而稀疏,能一次性求出所有的顶点到顶点的最短路径{int i, j, k;for (k = 1; k <= n; k++)for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)if (path[i][k] + path[k][j] < path[i][j])path[i][j] = path[i][k] + path[k][j];}void input(int line){int a, b, weight;for(int i=0; i<line; i++){scanf("%d%d%d", &a, &b, &weight);if(path[a][b] > weight) //有多条路,保存最短的那条{path[a][b] = weight;path[b][a] = weight; //无向图双向}}}int main(){int n, line;scanf("%d%d", &n, &line);initialize(n);input(line);spfa(n, 1);printf("%d\n\n", dist[n]);return 0;}
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原文地址:http://www.cnblogs.com/sober-reflection/p/83a93fc336fe2bf9cbbbb7090af2c932.html