PAT甲级专题|最短路
Posted fisherss
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了PAT甲级专题|最短路相关的知识,希望对你有一定的参考价值。
PAT甲级最短路
主要算法:dijkstra 求最短最长路、dfs图论搜索。
1018,dijkstra记录路径 + dfs搜索路径最值
25分,错误点暂时找不出。。
如果只用dijkstra没法做,只能得20分
#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 510;
int cmax,n,ter,m;
int caps[maxn];
int g[maxn][maxn];
int vis[maxn];
vector<int> pre[maxn];
int dist[maxn];
int half;
vector<int> paths;
vector<int> temp;
int minsend = inf;
int mintake = inf;
//dfs
void dfs(int x){
temp.push_back(x);
if(x == 0){
int need = 0;
int take = 0;
for(int i=0;i<temp.size();i++){
if(temp[i] == 0) continue;
if(caps[temp[i]] - half > 0){
int d = caps[temp[i]] - half;
if(need >= d) need -= d;
else{
need = 0;
take += (d - need);
}
}else{
need += (half - caps[temp[i]]);
}
}
if(need < minsend){
minsend = need;
mintake = take;
paths = temp;
}else if(need == minsend){
if(mintake > take){
mintake = take;
paths = temp;
}
}
temp.pop_back();
return;
}
for(int i=0;i<pre[x].size();i++){
dfs(pre[x][i]);
}
temp.pop_back();
}
void dijkstra(){
memset(vis,0,sizeof(vis));
memset(dist,inf,sizeof(dist));
int half = cmax/2;
dist[0] = 0;
for(int i=0;i<=n;i++){
int v,min_w = inf;
for(int j=0;j<=n;j++){
if(!vis[j] && dist[j] < min_w){
min_w = dist[j];
v = j;
}
}
if(min_w == inf) return;
vis[v] = 1;
for(int j=0;j<=n;j++){
if(vis[j] || g[v][j] == inf) continue;
if(g[v][j] + dist[v] < dist[j]){
dist[j] = g[v][j] + dist[v];
pre[j].clear();
pre[j].push_back(v);
}else if(g[v][j] + dist[v] == dist[j]){
pre[j].push_back(v);
}
}
}
}
int main(){
memset(g,inf,sizeof(g));
cin>>cmax>>n>>ter>>m;
half = cmax/2;
caps[0] = 0;
for(int i=1;i<=n;i++) cin>>caps[i];
for(int i=1;i<=m;i++){
int u,v,w;
cin>>u>>v>>w;
g[u][v] = w;
g[v][u] = w;
}
dijkstra();
dfs(ter);
cout<<minsend<<" ";
for(int i=paths.size()-1;i>0;i--){
cout<<paths[i]<<"->";
}
cout<<paths[0];
cout<<" "<<mintake<<endl;
return 0;
}
/*
10 3 3 5
6 7 10
0 1 1
0 2 1
0 3 2
1 3 1
2 3 1
*/
1030,多边权,多条更新
#include<bits/stdc++.h>
using namespace std;
/*
dijkstra:双边权
*/
const int inf = 0x3f3f3f3f;
const int maxn = 510;
int n,m,s,d;
int dist[maxn];
int path[maxn];
int cost[maxn];
int vis[maxn];
struct edge{
int v,w,c;
edge(int vv,int ww,int cc){
v = vv;
w = ww;
c = cc;
}
};
vector<int> paths;
vector<edge> g[maxn];
void dijkstra(int ss){
memset(vis,0,sizeof(vis));
memset(dist,inf,sizeof(dist));
memset(cost,inf,sizeof(cost));
dist[ss] = 0;
cost[ss] = 0;
path[ss] = -1;
for(int i=0;i<n;i++){
int v,min_w = inf;
for(int j=0;j<n;j++){
if(!vis[j] && dist[j] < min_w){
min_w = dist[j];
v = j;
}
}
if(min_w == inf) return;
vis[v] = 1;
for(int j=0;j<g[v].size();j++){
int x = g[v][j].v;
if(!vis[x] && dist[v] + g[v][j].w < dist[x]){
cost[x] = cost[v] + g[v][j].c;
dist[x] = dist[v] + g[v][j].w;
path[x] = v;
}else if(!vis[x] && dist[v] + g[v][j].w == dist[x]){
if(cost[x] > cost[v] + g[v][j].c){
path[x] = v;
cost[x] = cost[v] + g[v][j].c;
}
}
}
}
}
int main(){
cin>>n>>m>>s>>d;
for(int i=1;i<=m;i++){
int u,v,w,c;
cin>>u>>v>>w>>c;
g[u].push_back(edge(v,w,c));
g[v].push_back(edge(u,w,c));
}
dijkstra(s);
int cur = d;
while(cur != -1){
paths.push_back(cur);
cur = path[cur];
}
for(int i=paths.size()-1;i>=0;i--){
cout<<paths[i]<<" ";
}
cout<<dist[d]<<" "<<cost[d]<<endl;
return 0;
}
1087,记录所有路径,dfs搜索路径最值
#include<bits/stdc++.h>
using namespace std;
/*
map映射:string <-> cityId Name <-> Int
dijkstra:找出最短路的长度 以及所有最短路径(存入到pre容器中)
dfs:统计最短路的数量 从终点触发在出口判断更新所需的权值
*/
const int inf = 0x3f3f3f3f;
const int maxn = 210;
map<string, int> mp;
map<int,string> mp2;
int n,k;
string start;
int haps[maxn];
int g[maxn][maxn];
int nums;
int vis[maxn];
int dist[maxn];
int ter = 0;
vector<int> pre[maxn];
vector<int> temp,path;
int maxhap = 0;
double maxave = 0;
void dfs(int x){
temp.push_back(x);
if(x == n){
int curhap = 0;
double curave = 0;
for(int i=temp.size()-1;i>=0;i--){
curhap += haps[temp[i]];
}
if(temp.size() == 1){
curave = 0;
}else{
curave = curhap*1.0/(temp.size()-1);
}
if(maxhap < curhap){
maxhap = curhap;
maxave = curave;
path = temp;
}else if(maxhap == curhap){
if(maxave < curave){
maxave = curave;
path = temp;
}
}
nums++;
temp.pop_back();
return;
}
for(int i=0;i<pre[x].size();i++){
dfs(pre[x][i]);
}
temp.pop_back();
}
void dijkstra(int s){
memset(vis,0,sizeof(vis));
memset(dist,inf,sizeof(dist));
dist[s] = 0;
for(int i=1;i<=n;i++){
int v,min_w = inf;
for(int j=1;j<=n;j++){
if(!vis[j] && dist[j] < min_w){
v = j;
min_w = dist[j];
}
}
vis[v] = 1;
if(min_w == inf) return;
for(int j=1;j<=n;j++){
if(!vis[j] && g[v][j] != inf){
if(dist[j] > dist[v] + g[v][j]){
dist[j] = dist[v] + g[v][j];
pre[j].clear();
pre[j].push_back(v);
}else if(dist[j] == dist[v] + g[v][j]){
pre[j].push_back(v);
}
}
}
}
}
int main(){
memset(g,inf,sizeof(g));
cin>>n>>k>>start;
for(int i=1;i<=n-1;i++){
string city;
int hap;
cin>>city>>hap;
mp[city] = i;
mp2[i] = city;
haps[i] = hap;
}
mp[start] = n;
mp2[n] = start;
haps[n] = 0;
for(int i=1;i<=k;i++){
string c1,c2;
int cost;
cin>>c1>>c2>>cost;
int u = mp[c1];
int v = mp[c2];
g[u][v] = cost;
g[v][u] = cost;
}
ter = mp["ROM"];
dijkstra(n);
dfs(ter);
cout<<nums<<" "<<dist[ter]<<" "<<maxhap<<" "<<int(maxave)<<endl;
cout<<start;
for(int i=path.size()-2;i>0;i--){
cout<<"->"<<mp2[path[i]];
}
cout<<"->ROM"<<endl;
return 0;
}
1111,多边权、记录路径、多条件更新
#include<bits/stdc++.h>
using namespace std;
/*
18分
*/
const int inf = 0x3f3f3f3f;
const int maxn = 510;
int source,ter;
int n,m;
int vis1[maxn];
int vis2[maxn];
struct edge{
int v;
int length;
int time;
edge(int vv,int len,int ti){
v = vv;
length = len;
time = ti;
}
};
vector<edge> g[maxn];
int dist1[maxn];
int time1[maxn];
int time2[maxn];
int path1[maxn];
int path2[maxn];
int nums[maxn];
vector<int> ans1;
vector<int> ans2;
void dijkstra1(){
memset(vis1,0,sizeof(vis1));
memset(dist1,inf,sizeof(dist1));
memset(time1,inf,sizeof(time1));
dist1[source] = 0;
time1[source] = 0;
path1[source] = -1;
for(int i=0;i<n;i++){
int v,min_w = inf;
for(int j=0;j<n;j++){
if(!vis1[j] && dist1[j] < min_w){
v = j;
min_w = dist1[j];
}
}
vis1[v] = 1;
if(min_w == inf) return;
for(int j=0;j<g[v].size();j++){
int u = g[v][j].v;
if(!vis1[u]){
if(dist1[u] > dist1[v] + g[v][j].length){
time1[u] = time1[v] + g[v][j].time;
dist1[u] = dist1[v] + g[v][j].length;
path1[u] = v;
}else if(dist1[u] == dist1[v] + g[v][j].length && time1[u] > time1[v] + g[v][j].time){
time1[u] = time1[v] + g[v][j].time;
path1[u] = v;
}
}
}
}
}
void dijkstra2(){
memset(vis2,0,sizeof(vis2));
memset(nums,inf,sizeof(nums));
memset(time2,inf,sizeof(time2));
nums[source] = 1;
time2[source] = 0;
path2[source] = -1;
for(int i=0;i<n;i++){
int v,min_w = inf;
for(int j=0;j<n;j++){
if(!vis2[j] && time2[j] < min_w){
v = j;
min_w = time2[j];
}
}
vis2[v] = 1;
if(min_w == inf) return;
for(int j=0;j<g[v].size();j++){
int u = g[v][j].v;
if(!vis2[u]){
if(time2[u] > time2[v] + g[v][j].time){
time2[u] = time2[v] + g[v][j].time;
nums[u] = nums[v] + 1;
path2[u] = v;
}else if(time2[u] == time2[v] + g[v][j].time){
if(nums[u] >= nums[v] + 1){
path2[u] = v;
nums[u] = nums[v] + 1;
}
}
}
}
}
}
bool identical(){
int cur = ter;
while(cur != -1){
ans1.push_back(cur);
cur = path1[cur];
}
cur = ter;
while(cur != -1){
ans2.push_back(cur);
cur = path2[cur];
}
reverse(ans1.begin(),ans1.end());
reverse(ans2.begin(),ans2.end());
if(ans1.size() != ans2.size()) return false;
for(int i=0;i<ans1.size();i++){
if(ans1[i] != ans2[i])return false;
}
return true;
}
int main(){
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
cin>>n>>m;
for(int i=1;i<=m;i++){
int v1,v2,one,length,time;
cin>>v1>>v2>>one>>length>>time;
if(one == 0){
g[v1].push_back(edge(v2,length,time));
g[v2].push_back(edge(v1,length,time));
}else{
g[v1].push_back(edge(v2,length,time));
}
}
cin>>source>>ter;
dijkstra1();
dijkstra2();
if(identical()){
printf("Distance = %d; Time = %d: %d",dist1[ter],time2[ter],source);
for(int i=1;i<ans1.size();i++){
printf(" -> %d",ans1[i]);
}
printf("
");
}else{
printf("Distance = %d: %d",dist1[ter],source);
for(int i=1;i<ans1.size();i++){
printf(" -> %d",ans1[i]);
}
printf("
");
printf("Time = %d: %d",time2[ter],source);
for(int i=1;i<ans2.size();i++){
printf(" -> %d",ans2[i]);
}
printf("
");
}
return 0;
}
/*
4 4
0 1 1 1 2
0 2 1 2 1
1 3 1 2 2
2 3 1 1 1
0 3
*/
以上是关于PAT甲级专题|最短路的主要内容,如果未能解决你的问题,请参考以下文章
PAT甲级 All Roads Lead to Rome (dijkstra+dfs回溯)