题目1008：最短路径问题

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我是靠谱客的博主清秀大神，这篇文章主要介绍题目1008：最短路径问题，现在分享给大家，希望可以做个参考。

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：

复制代码

样例输出：

复制代码

1
9 11

注：注意重边，考虑当已确定集到未确定集中距离相等时，注意比较花费

题目比较简单用邻接矩阵+Dijkstra即可

复制代码

#include <iostream>
#include <stdio.h>
using namespace std;
int data[1001][1001][2];
int dis[1001];
int fare[1001];
int points[1001];
bool isOver(int points[],int n){
bool over = true;
for(int i = 1; i <= n; i++){
if(points[i] == 0){
over = false;
break;
}
}
return over;
}
int main(){
int n,m;
scanf("%d %d",&n,&m);
while(n != 0 && m != 0){
for(int i = 0; i < 1001;i++){//初始化
for(int j = 0; j < 1001;j++){
data[i][j][0] = -1;
data[i][j][1] = -1;
}
dis[i] = -1;
fare[i] = -1;
points[i] = 0;
}
for(int i = 0; i < m; i++){//读数据
int point1,point2,distance,money;
scanf("%d %d %d %d",&point1,&point2,&distance,&money);
data[point1][point2][0] = distance;
data[point1][point2][1] = money;
data[point2][point1][0] = distance;
data[point2][point1][1] = money;
}
int beginp,endp;
scanf("%d %d",&beginp,&endp);
dis[beginp] = 0;
fare[beginp] = 0;
points[beginp] = 1;
for(int i = 1; i <= n; i++){
if(data[beginp][i][0] != -1){
dis[i] = data[beginp][i][0];
fare[i] = data[beginp][i][1];
}
}
while(!isOver(points,n)){
int point = -1,anotherP= -1,minMoney,minDis = -1;
for(int i = 1; i <= n;i++){
if(points[i]==1){//已确定点
for(int j = 1; j <= n; j++){
if(points[j] == 0){//未确定点
if(data[i][j][0] ==-1)
continue;
if(minDis == -1 || minDis > data[i][j][0]){
minDis = data[i][j][0];
minMoney = data[i][j][1]+fare[i];
point = j;
anotherP = i;
}else if(minDis == data[i][j][0] && (fare[anotherP] + data[anotherP][point][1] > fare[i] + data[i][j][1])){//距离相同比较花费
minDis = data[i][j][0];
minMoney = data[i][j][1];
point = j;
anotherP = i;
}
}
}
}
}
//
points[point] = 1;
dis[point] = dis[anotherP]+ data[anotherP][point][0];
fare[point] = fare[anotherP] + data[anotherP][point][1];
for(int i = 1; i <= n ;i++){
if(points[i] == 0 &&data[point][i][0] != -1){
if(dis[i] == -1 || dis[i] > dis[point] +data[point][i][0]){
dis[i] = dis[point] + data[point][i][0];
}
}
}
}
printf("%d %dn",dis[endp],fare[endp]);
scanf("%d %d",&n,&m);
}
return 0;
}

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#include <iostream>
#include <stdio.h>
using namespace std;
int data[1001][1001][2];
int dis[1001];
int fare[1001];
int points[1001];
bool isOver(int points[],int n){
bool over = true;
for(int i = 1; i <= n; i++){
if(points[i] == 0){
over = false;
break;
}
}
return over;
}
int main(){
int n,m;
scanf("%d %d",&n,&m);
while(n != 0 && m != 0){
for(int i = 0; i < 1001;i++){//初始化
for(int j = 0; j < 1001;j++){
data[i][j][0] = -1;
data[i][j][1] = -1;
}
dis[i] = -1;
fare[i] = -1;
points[i] = 0;
}
for(int i = 0; i < m; i++){//读数据
int point1,point2,distance,money;
scanf("%d %d %d %d",&point1,&point2,&distance,&money);
data[point1][point2][0] = distance;
data[point1][point2][1] = money;
data[point2][point1][0] = distance;
data[point2][point1][1] = money;
}
int beginp,endp;
scanf("%d %d",&beginp,&endp);
dis[beginp] = 0;
fare[beginp] = 0;
points[beginp] = 1;
for(int i = 1; i <= n; i++){
if(data[beginp][i][0] != -1){
dis[i] = data[beginp][i][0];
fare[i] = data[beginp][i][1];
}
}
while(!isOver(points,n)){
int point = -1,anotherP= -1,minMoney,minDis = -1;
for(int i = 1; i <= n;i++){
if(points[i]==1){//已确定点
for(int j = 1; j <= n; j++){
if(points[j] == 0){//未确定点
if(data[i][j][0] ==-1)
continue;
if(minDis == -1 || minDis > data[i][j][0]){
minDis = data[i][j][0];
minMoney = data[i][j][1]+fare[i];
point = j;
anotherP = i;
}else if(minDis == data[i][j][0] && (fare[anotherP] + data[anotherP][point][1] > fare[i] + data[i][j][1])){//距离相同比较花费
minDis = data[i][j][0];
minMoney = data[i][j][1];
point = j;
anotherP = i;
}
}
}
}
}
//
points[point] = 1;
dis[point] = dis[anotherP]+ data[anotherP][point][0];
fare[point] = fare[anotherP] + data[anotherP][point][1];
for(int i = 1; i <= n ;i++){
if(points[i] == 0 &&data[point][i][0] != -1){
if(dis[i] == -1 || dis[i] > dis[point] +data[point][i][0]){
dis[i] = dis[point] + data[point][i][0];
}
}
}
}
printf("%d %dn",dis[endp],fare[endp]);
scanf("%d %d",&n,&m);
}
return 0;
}