畅通工程续
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23439 Accepted Submission(s): 8253
Problem Description
某省自从实行了很多年的畅通工程计划后,终于修建了很多路。不过路多了也不好,每次要从一个城镇到另一个城镇时,都有许多种道路方案可以选择,而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。
现在,已知起点和终点,请你计算出要从起点到终点,最短需要行走多少距离。
Input
本题目包含多组数据,请处理到文件结束。
每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000),分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0~N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N),分别代表起点和终点。
Output
对于每组数据,请在一行里输出最短需要行走的距离。如果不存在从S到T的路线,就输出-1.
Sample Input
3 3
0 1 1
0 2 3
1 2 1
0 2
3 1
0 1 1
1 2
Sample Output
Dijistra (15ms):
#include <stdio.h>
#include <string.h>
#include <vector>
#define N 220
using namespace std;
struct Road{
int next;
int dis;
};
vector<Road>City[N];
int CityNum, RoadNum;
int dij[N];
bool vis[N];
void Init(){
memset(vis, 0, sizeof(vis));
for(int i = 0; i <= CityNum; i ++){
City[i].clear();
dij[i] = 0xfffffff;
}
}
void Dijistra(int cur){
dij[cur] = 0;
while(vis[cur] == 0){
vis[cur] = 1;
for(int i = 0; i < City[cur].size(); i ++){
int next = City[cur].at(i).next;
int dis = City[cur].at(i).dis;
if(dij[next] > dij[cur] + dis){
dij[next] = dij[cur] + dis;
}
}
int min = 0xfffffff;
for(int i = 0; i < CityNum; i ++){
if(!vis[i] && min > dij[i]){
min = dij[i];
cur = i;
}
}
}
}
int main()
{
Road nw;
while(scanf("%d%d", &CityNum, &RoadNum) != EOF){
Init();
int start, end, distance;
for(int i = 0; i < RoadNum; i ++){
scanf("%d%d%d", &start, &end, &distance);
nw.next = start;
nw.dis = distance;
City[end].push_back(nw);
nw.next = end;
City[start].push_back(nw);
}
scanf("%d%d", &start, &end);
Dijistra(start);
if(dij[end] == 0xfffffff){
printf("-1\n");
}
else{
printf("%d\n", dij[end]);
}
}
return 0;
}
Floyd (78ms):
#include <stdio.h>
#include <algorithm>
#define N 205
using namespace std;
int map[N][N];
int CityNum, RoadNum;
int floyd(int start, int end){
for(int k = 0; k < CityNum; k ++){
for(int i = 0; i < CityNum; i ++){
for(int j = 0; j < CityNum; j ++){
map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
}
}
}
return map[start][end] < 0xfffffff ? map[start][end] : -1;
}
int main()
{
int start, end, len;
while(scanf("%d%d", &CityNum, &RoadNum) != EOF){
for(int i = 0; i < CityNum; i ++){
for(int j = 0; j < CityNum; j ++){
map[i][j] = (i == j ? 0 : 0xfffffff);
}
}
for(int i = 0; i < RoadNum; i ++){
scanf("%d%d%d", &start, &end, &len);
if(len < map[start][end]){ // 注使用邻接矩阵时 需判断重连
map[end][start] = map[start][end] = len; // 无向图
}
}
scanf("%d%d", &start, &end);
int ans = floyd(start, end);
printf("%d\n", ans);
}
return 0;
}
Bellman_Ford(15ms):
#include <stdio.h>
struct ROAD{
int from;
int to;
int len;
};
ROAD Road[2020];
int CityNum, RoadNum;
int dis[220];
void Bellman_Ford(int start){
for(int i = 0; i < CityNum; i ++){
dis[i] = 0xfffffff;
}
dis[start] = 0;
for(int i = 0; i < CityNum; i ++){
for(int j = 0; j < RoadNum * 2; j ++){
if(dis[Road[j].to] > dis[Road[j].from] + Road[j].len){
dis[Road[j].to] = dis[Road[j].from] + Road[j].len;
}
}
}
}
int main()
{
int start, end;
while(scanf("%d%d", &CityNum, &RoadNum) != EOF){
for(int i = 0; i < RoadNum; i ++){
scanf("%d%d%d", &Road[i].from, &Road[i].to, &Road[i].len);
Road[RoadNum + i].from = Road[i].to;
Road[RoadNum + i].to = Road[i].from;
Road[RoadNum + i].len = Road[i].len;
}
scanf("%d%d", &start, &end);
Bellman_Ford(start);
if(dis[end] == 0xfffffff){
printf("-1\n");
}
else{
printf("%d\n", dis[end]);
}
}
return 0;
}
SPFA (0ms):
#include <stdio.h>
#include <string.h>
#include <queue>
#define N 220
using namespace std;
int CityNum, RoadNum;
int map[N][N], dis[N];
bool vis[N];
void Init(){
memset(vis, 0, sizeof(vis));
for(int i = 0; i < CityNum; i ++){
dis[i] = 0xfffffff;
for(int j = 0; j < CityNum; j ++){
map[i][j] = 0xfffffff;
}
}
}
void SPFA(int start){
queue<int>q;
q.push(start);
dis[start] = 0;
while(!q.empty()){
int cur = q.front();
q.pop();
vis[cur] = 0;
for(int i = 0; i < CityNum; i ++){
if(dis[i] > map[cur][i] + dis[cur]){
dis[i] = map[cur][i] + dis[cur];
if(!vis[i]){
q.push(i);
vis[i] = 1;
}
}
}
}
}
int main()
{
int start, end, len;
while(scanf("%d%d", &CityNum, & RoadNum) != EOF){
Init();
for(int i = 0; i < RoadNum; i ++){
scanf("%d%d%d", &start, &end, &len);
if(map[start][end] > len){
map[start][end] = map[end][start] = len;
}
}
scanf("%d%d", &start, &end);
SPFA(start);
if(dis[end] == 0xfffffff){
printf("-1\n");
}
else{
printf("%d\n", dis[end]);
}
}
return 0;
}