hdu 1385 Minimum Transport Cost
点击打开链接hdu 1385
思路：最短路+SPFA+路径记录+路径字典序比较
分析：
1 题目要求的单源的最短路，所以可以选择任意一种单源最短路来求解
2 题目还要求在路径和相同情况下要字典序小的，那么就要在更新dis数组的时候进行更新路径，如果是dis[i]>tmp,那么直接更新；如果是dis[i] == tmp的情况下，那么就要求出star->i 和 star->x的路径进行比较，然后判断能否更新.
3 注意询问的时候可能问的是同一个点。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define MAXN 110
#define INF 0xFFFFFFF
int n , star , end , pos;
int value[MAXN][MAXN];
int tmp_value[MAXN][MAXN];
int dis[MAXN];
int charge[MAXN];
int vis[MAXN];
int route[MAXN];
int father[MAXN];
int tmp_charge[MAXN];
queue<int>q;
/*初始化数组*/
void init(){
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++)
value[i][j] = INF;
value[i][i] = 0;
}
}
/*dfs找路径*/
void dfs(int num , char *str){
if(num == -1)
return;
dfs(father[num] , str);
str[pos++] = num+'0';
}
void SPFA(int s){
memset(vis , 0 , sizeof(vis));
for(int i = 1 ; i <= n ; i++){
dis[i] = INF;
father[i] = -1;
}
vis[s] = 1;
dis[s] = 0;
q.push(s);
while(!q.empty()){
int x = q.front();
q.pop();
vis[x] = 0;
for(int i = 1 ; i <= n ; i++){
int tmp = dis[x] + tmp_value[x][i] + tmp_charge[x];
/*dis[i]>tmp直接更新*/
if(dis[i] > tmp){
dis[i] = tmp;
father[i] = x;
if(!vis[i]){
vis[i] = 1;
q.push(i);
}
}
/*dis[i] == tmp的时候求出路径*/
else if(dis[i] == tmp && x != i && dis[i] != INF){
char ch1[MAXN] = {'\0'} , ch2[MAXN] = {'\0'};
pos = 0;
dfs(i , ch1);
pos = 0;
dfs(x , ch2);
ch2[pos] = x+'0';/*这个地方注意就是x要加上去*/
/*如果字典序小就要更新*/
if(strcmp(ch1 , ch2) > 0){
father[i] = x;
if(!vis[i]){
vis[i] = 1;
q.push(i);
}
}
}
}
}
}
int main(){
int x , y , num , cnt , ans;
while(scanf("%d" , &n) && n){
init();
/*输入边*/
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++){
scanf("%d" , &num);
if(num != -1 && value[i][j] > num)
value[i][j] = num;
}
}
/*输入过路费*/
for(int i = 1 ; i <= n ; i++)
scanf("%d" , &charge[i]);
/*询问*/
while(1){
scanf("%d%d" , &star , &end);
if(star == -1 && end == -1)
break;
/*数组复制*/
memcpy(tmp_charge , charge , sizeof(charge));
memcpy(tmp_value , value , sizeof(value));
tmp_charge[star] = tmp_charge[end] = 0;/*起点和终点的过路费为0*/
SPFA(star);
/*输出*/
printf("From %d to %d :\n" , star , end);
printf("Path: ");
memset(route , 0 , sizeof(route));
ans = dis[end];
if(star != end){
cnt = 0;
x = end;
while(1){
y = father[x];
route[cnt++] = y;
if(y == star)
break;
x = y;
}
for(int i = cnt-1 ; i >= 0 ; i--)
printf("%d-->" , route[i]);
}
printf("%d\n" , end);
printf("Total cost : %d\n\n" , ans);
}
}
return 0;
}
2 floyd ，由于floyd是每一次都用k去更新dis，所以我们只要在更新的同时更新path即可，由于k是从1-n的，那么这样最后的path肯定是字典序最小的。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 110
#define INF 0xFFFFFFF
int n;
int value[MAXN][MAXN];
int charge[MAXN];
int path[MAXN][MAXN];
void init(){
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++){
value[i][j] = INF;
path[i][j] = j;
}
value[i][i] = 0;
}
}
void floyd(){
for(int k = 1 ; k <= n ; k++){
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++){
int tmp = value[i][k]+value[k][j]+charge[k];
if(value[i][j] > tmp){
value[i][j] = tmp;
path[i][j] = path[i][k];
}
else if(value[i][j] == tmp && path[i][j] > path[i][k])
path[i][j] = path[i][k];
}
}
}
}
int main(){
int x , y , num , star , end;
while(scanf("%d" , &n) && n){
init();
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++){
scanf("%d" , &num);
if(num != -1)
value[i][j] = num;
}
}
for(int i = 1 ; i <= n ; i++)
scanf("%d" , &charge[i]);
floyd();
while(1){
scanf("%d%d" , &star , &end);
if(star == -1 && end == -1)
break;
printf("From %d to %d :\n" , star , end);
printf("Path: %d" , star);
x = star , y = end;
if(star != end){
while(1){
int p = path[x][y];
printf("-->%d" , p);
if(p == end)
break;
x = p;
}
}
printf("\n");
printf("Total cost : %d\n\n" , value[star][end]);
}
}
return 0;
}