POJ2457 dijksta求最短路&&输出走过的路径

http://poj.org/problem?id=2457

题意：求起点1到终点m的最短距离。这个可以spfa||dijkstra。这题关键的是要输出从1--m走过的路径。用path[]记录下，每一步的祖先，用shortest[]逆序输出即可。这题re了n次。到现在也不理解，为什么先

if(dis[m]==INF){//为什么先判断就不会re.不明白
printf("-1\n");
continue;
}

这个就没问题。




View Code

// I'm lanjiangzhou
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

#define maxn 2100
int n,m;
int edges[maxn][maxn];
int S[maxn];
int dis[maxn];
int path[maxn];
int shortest[maxn];

void dijkstra(int v0){
    for(int i=1;i<=m;i++){
        dis[i]=edges[v0][i];
        S[i]=0;
        if(i!=v0&&dis[i]<INF) path[i]=v0;
        else path[i]=-1;
    }

    S[v0]=1;  dis[v0]=0;
    for(int i=1;i<=m-1;i++){
        int min=INF,u=v0;
        for(int j=1;j<=m;j++){
            if(!S[j]&&dis[j]<min){
                u=j;
                min=dis[j];
            }
        }
        S[u]=1;
        for(int k=1;k<=m;k++){
            if(!S[k]&&edges[u][k]<INF&&dis[u]+edges[u][k]<dis[k]){
                dis[k]=dis[u]+edges[u][k];
                path[k]=u;
            }
        }
    }
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        int u,v;
        //³õʼ»¯
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                edges[i][j]=INF;
            }
            edges[i][i]=0;
        }
        for(int i=1;i<=n;i++){
            scanf("%d%d",&u,&v);
            edges[u][v]=1;
        }
        dijkstra(1);
        if(dis[m]==INF){//为什么先判断就不会re.不明白
            printf("-1\n");
            continue;
        }
        //printf("dis[m]=%d\n",dis[m]);
        memset(shortest,0,sizeof(shortest));
        int k=0;
        shortest[k]=m;
        while(path[shortest[k]]!=1){
            k++;  shortest[k]=path[shortest[k-1]];
        }
        k++;  shortest[k]=1;
//        if(dis[m]==INF){//为什么把dis[m]==INF放在这里就不会re.擦，re了13次。
//            printf("-1\n");
//        }
       // else {
            printf("%d\n",dis[m]+1);
            for(int j=k;j>0;j--){
                printf("%d\n",shortest[j]);
            }
            printf("%d\n",shortest[0]);
        //}
    }
    return 0;
}