九度OJ 1447 最短路 1008 最短路径问题

题目地址：http://ac.jobdu.com/problem.php?pid=1447

题目描述：

在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？

输入：

输入包括多组数据。每组数据第一行是两个整数N、M（N<=100，M<=10000），N表示成都的大街上有几个路口，标号为1的路口是商店所在地，标号为N的路口是赛场所在地，M则表示在成都有几条路。N=M=0表示输入结束。接下来M行，每行包括3个整数A，B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A与路口B之间有一条路，我们的工作人员需要C分钟的时间走过这条路。输入保证至少存在1条商店到赛场的路线。
当输入为两个0时，输入结束。

输出：

对于每组输入，输出一行，表示工作人员从商店走到赛场的最短时间。

样例输入：

2 1
1 2 3
3 3
1 2 5
2 3 5
3 1 2
0 0

样例输出：

3
2

Floyd算法：

#include <stdio.h>
#include <limits.h>
 
void Init (int graph[101][101], int N){
    int i;
    int j;
    for (i=1; i<=N; ++i){
        for (j=1; j<=N; ++j){
            graph[i][j] = INT_MAX;
        }
    }
    for (i=1; i<=N; ++i)
        graph[i][i] = 0;
}
 
void Floyd (int graph[101][101], int N){
    int i;
    int j;
    int k;
    for (k=1; k<=N; ++k){
        for (i=1; i<=N; ++i){
            for (j=1; j<=N; ++j){
                if (graph[i][k] == INT_MAX || graph[k][j] == INT_MAX)
                    continue;
                if (graph[i][j] > graph[i][k] + graph[k][j])
                    graph[i][j] =graph[i][k] + graph[k][j];
            }
        }
    }
}
 
int main(void){
    int N;
    int M;
    int graph[101][101];
    int start;
    int end;
    int length;
 
    while (scanf ("%d%d", &N, &M) != EOF && N != 0 && M != 0){
        Init (graph, N);
        while (M-- != 0){
            scanf ("%d%d%d", &start, &end, &length);
            graph[start][end] = length;
            graph[end][start] = length;
        }
        Floyd (graph, N);
        printf ("%d
", graph[1][N]);
    }
 
 
    return 0;
}

题目地址：http://ac.jobdu.com/problem.php?pid=1008

题目描述：

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

输入：

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点t。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

输出：

输出 一行有两个数， 最短距离及其花费。

样例输入：

3 2
1 2 5 6
2 3 4 5
1 3
0 0

样例输出：

9 11

来源：

2010年浙江大学计算机及软件工程研究生机试真题

Dijkstra算法：

1、邻接矩阵：

#include <stdio.h>
#include <limits.h>
 
#define MAXN 1001
 
typedef struct road{
    int len;
    int cost;
}Road;
 
void Init (Road graph[MAXN][MAXN], Road minlength[MAXN],int n){
    int i, j;
    for (i=1; i<=n; ++i){
        for (j=1; j<=n; ++j){
            graph[i][j].len = INT_MAX;
            graph[i][j].cost = INT_MAX;
        }
    }
    for (i=1; i<=n; ++i){
        graph[i][i].len = 0;
        graph[i][i].cost = 0;
        minlength[i].len = INT_MAX;
        minlength[i].cost = INT_MAX;
    }
}
 
void Dijkstra (Road graph[MAXN][MAXN], Road minlength[MAXN], int N, int s, int t){
    int visited[MAXN];
    int i;
    int node;
    int reminder;
    int minlen;
    int mincost;
    int index;
  
    for (i=1; i<=N; ++i)
        visited[i] = 0;
    visited[s] = 1;
    minlength[s].len = 0;
    minlength[s].cost = 0;
    reminder = N - 1;
    node = s;
    while (reminder-- != 0){
        for (i=1; i<=N; ++i){
            if (graph[node][i].len == INT_MAX || graph[node][i].len == 0)
                continue;
            else{
                if ((minlength[node].len + graph[node][i].len < minlength[i].len) ||
                    ((minlength[node].len + graph[node][i].len == minlength[i].len) &&
                    (minlength[node].cost + graph[node][i].cost < minlength[i].cost))){
                    minlength[i].len = minlength[node].len + graph[node][i].len;
                    minlength[i].cost = minlength[node].cost + graph[node][i].cost;
                }
            }
        }
        minlen = INT_MAX;
        mincost = INT_MAX;
        for (i=1; i<=N; ++i){
            if (visited[i] == 0 ){
                if (minlength[i].len < minlen || (minlength[i].len == minlen &&
                    minlength[i].cost < mincost)){
                        minlen = minlength[i].len;
                        mincost = minlength[i].cost;
                        index = i;
                }
            }
        }
        visited[index] = 1;
        node = index;
    }
}
 
int main(void){
    int n, m;
    int a, b, d, p;
    int s, t;
    Road graph[MAXN][MAXN];
    Road minlength[MAXN];
 
    while (scanf ("%d%d", &n, &m) != EOF && n != 0 && m != 0){
        Init (graph, minlength, n);
        while (m-- != 0){
            scanf ("%d%d%d%d", &a, &b, &d, &p);
            graph[a][b].len = d;
            graph[a][b].cost = p;
            graph[b][a].len = d;
            graph[b][a].cost = p;
        }
        scanf ("%d%d", &s, &t);
        Dijkstra (graph, minlength, n, s, t);
        printf ("%d %d
", minlength[t].len, minlength[t].cost);
    }
 
    return 0;
}

2、邻接表：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
 
#define MAXN 1001
 
typedef struct edge{
    int end;
    int len;
    int cost;
    struct edge * next;
}Edge;
 
typedef struct vertex{
    int start;
    struct edge * next;
}Vertex;
 
typedef struct road{
    int len;
    int cost;
}Road;
 
void Init (Vertex graph[MAXN], Road minlength[MAXN], int n){
    int i;
    for (i=1; i<=n; ++i){
        graph[i].start = i;
        graph[i].next = NULL;
        minlength[i].len = INT_MAX;
        minlength[i].cost = INT_MAX;
    }
}
 
void Insert (Vertex graph[MAXN], int start, int end, int len, int cost){
    Edge * t = graph[start].next;
    Edge * pre = NULL;
    Edge * p = NULL;
    while (t != NULL && t->end != end){
        pre = t;
        t = t->next;
    }
    if (t == NULL){
        p = (Edge *)malloc(sizeof(Edge));
        p->end = end;
        p->len = len;
        p->cost = cost;
        p->next = NULL;
        if (pre != NULL)
            pre->next = p;
        else
            graph[start].next = p;
    }
    else{
        if (len < t->len || (len == t->len && cost < t->cost)){
            t->len = len;
            t->cost = cost;
        }
    }
}
 
void Dijkstra (Vertex graph[MAXN], Road minlength[MAXN], int N, int start, int end){
    int visited[MAXN];
    int i;
    int node;
    int reminder;
    int minlen;
    int mincost;
    int index;
    Edge * p = NULL;
    Edge * q = NULL;
  
    for (i=1; i<=N; ++i)
        visited[i] = 0;
    visited[start] = 1;
    minlength[start].len = 0;
    minlength[start].cost = 0;
    reminder = N - 1;
    node = start;
    while (reminder-- != 0 && node != end){
        p = graph[node].next;
        while (p != NULL){
            if ((p->len + minlength[node].len < minlength[p->end].len) ||
                ((p->len + minlength[node].len == minlength[p->end].len) &&
                (p->cost + minlength[node].cost < minlength[p->end].cost))){
                    minlength[p->end].len = p->len + minlength[node].len;
                    minlength[p->end].cost = p->cost + minlength[node].cost;
            }
            p = p->next;
        }
        minlen = INT_MAX;
        mincost = INT_MAX;
        for (i=1; i<=N; ++i){
            if (visited[i] == 0 ){
                if (minlength[i].len < minlen || (minlength[i].len == minlen &&
                    minlength[i].cost < mincost)){
                        minlen = minlength[i].len;
                        mincost = minlength[i].cost;
                        index = i;
                }
            }
        }
        visited[index] = 1;
        node = index;
    }
}
 
int main(void){
    int n, m;
    int a, b, d, p;
    int s, t;
    Vertex graph[MAXN];
    Road minlength[MAXN];
 
    while (scanf ("%d%d", &n, &m) != EOF && n != 0 && m != 0){
        Init (graph, minlength, n);
        while (m-- != 0){
            scanf ("%d%d%d%d", &a, &b, &d, &p);
            Insert (graph, a, b, d, p);
            Insert (graph, b, a, d, p);
        }
        scanf ("%d%d", &s, &t);
        Dijkstra (graph, minlength, n, s, t);
        printf ("%d %d
", minlength[t].len, minlength[t].cost);
    }
 
    return 0;
}

参考资料：维基百科 -- Floyd-Warshall算法

                 维基百科 -- Dijkstra算法