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\n", 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
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\n", 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\n", minlength[t].len, minlength[t].cost);
}
return 0;
}
参考资料： 维基百科 -- Floyd-Warshall算法