http://acm.hdu.edu.cn/showproblem.php?
pid=1385
求最短路。要求输出字典序最小的路径。
spfa：拿一个pre[]记录前驱，不同的是在松弛的时候。要考虑和当前点的dis值相等的情况，解决的办法是dfs找出两条路径中字典序较小的。pre[]去更新。
把路径当做字符串处理。
我仅仅用之前的pre去更新当前点，并没考虑到起点到当前点的整个路径，事实上这样并不能保证是字典序最小。wa了N次。于是乎搜了下题解，发现用spfa解的非常少。看到了某大牛的解法如上，感觉非常赞的想法。
#include <stdio.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 110;
int n;
int tax[maxn];
int Map[maxn][maxn];
int dis[maxn],inque[maxn];
int pre[maxn];
int pos = 0;
void dfs(int u, char *s)
{
if(u == -1)
return;
dfs(pre[u],s);
s[pos++] = u+'0';
}
bool solve(int v, int u)
{
char s1[maxn],s2[maxn];
//寻找先前的路径
pos = 0;
dfs(v,s1);
s1[pos] = ' ';
//寻找由u更新的路径
pos = 0;
dfs(u,s2);
s2[pos++] = v+'0';
s2[pos] = ' ';
if(strcmp(s1,s2) > 0) return true;
return false;
}
int spfa(int s, int t)
{
queue <int> que;
memset(dis,INF,sizeof(dis));
memset(inque,0,sizeof(inque));
memset(pre,-1,sizeof(pre));
dis[s] = 0;
inque[s] = 1;
que.push(s);
while(!que.empty())
{
int u = que.front();
que.pop();
inque[u] = 0;
for(int v = 1; v <= n; v++)
{
if(Map[u][v] > 0)
{
int tmp = dis[u] + Map[u][v] + tax[v];
if(tmp < dis[v])//直接更新
{
dis[v] = tmp;
pre[v] = u;
if(!inque[v])
{
inque[v] = 1;
que.push(v);
}
}
else if(tmp == dis[v] && solve(v,u))
{
pre[v] = u;
}
}
}
}
return dis[t];
}
void output(int s, int t)
{
int res[maxn];
int cnt = 0;
int tmp = t;
while(pre[tmp] != -1)
{
res[cnt++] = tmp;
tmp = pre[tmp];
}
res[cnt] = s;
printf("Path: ");
for(int i = cnt; i >= 1; i--)
printf("%d-->",res[i]);
printf("%d
",res[0]);
}
int main()
{
while(~scanf("%d",&n) && n)
{
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
scanf("%d",&Map[i][j]);
}
for(int i = 1; i <= n; i++)
scanf("%d",&tax[i]);
int u,v;
while(~scanf("%d %d",&u,&v))
{
if(u == -1 && v == -1) break;
int tmp1 = tax[u];//备份
int tmp2 = tax[v];
tax[u] = 0;
tax[v] = 0;
printf("From %d to %d :
",u,v);
int ans = spfa(u,v);
output(u,v);
printf("Total cost : %d
",ans);
//恢复备份
tax[u] = tmp1;
tax[v] = tmp2;
}
}
return 0;
}
floyd:pre[i][j]记录i到j路径上距离i近期的点，输出路径时按pre正向输出。感觉floyd非常强大啊，还能够这样记录路径。
#include <stdio.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <stack>
#include <queue>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 110;
int Map[maxn][maxn];
int tax[maxn];
int pre[maxn][maxn];
int n;
void floyd()
{
//初始化
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
pre[i][j] = j;
for(int k = 1; k <= n; k++)
{
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(Map[i][k] > 0 && Map[k][j] > 0)
{
int tmp = Map[i][k] + Map[k][j] + tax[k];
if(tmp < Map[i][j]) //小于直接更新
{
Map[i][j] = tmp;
pre[i][j] = pre[i][k];
}
else if(tmp == Map[i][j])
{
pre[i][j] = min(pre[i][k],pre[i][j]);
}
}
}
}
}
}
void output(int s, int t)
{
printf("Path: ");
int tmp = s;
while(tmp != t)
{
printf("%d-->",tmp);
tmp = pre[tmp][t];
}
printf("%d
",t);
}
int main()
{
while(~scanf("%d",&n) && n)
{
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
{
scanf("%d",&Map[i][j]);
if(Map[i][j] == -1)
Map[i][j] = INF;
}
for(int i = 1; i <= n; i++)
scanf("%d",&tax[i]);
int u,v;
floyd();
while(~scanf("%d %d",&u,&v))
{
if(u == -1 && v == -1) break;
printf("From %d to %d :
",u,v);
output(u,v);
printf("Total cost : %d
",Map[u][v]);
}
}
return 0;
}