POJ 3255 Roadblocks (次短路)

题意：给定一个图，求一条1-n的次短路。

析：次短路就是最短路再长一点呗，我们可以和求最短路一样，再多维护一个数组，来记录次短路。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

vector<int> G[maxn], w[maxn];
int d1[maxn], d2[maxn];

int dijkstra(){
  priority_queue<P, vector<P>, greater<P> > pq;
  pq.push(P(0, 1));
  memset(d1, INF, sizeof d1);
  memset(d2, INF, sizeof d2);
  d1[1] = 0;

  while(!pq.empty()){
    P p = pq.top();  pq.pop();
    int v = p.second, d = p.first;
    if(d2[v] < d)  continue;
    for(int i = 0; i < G[v].size(); ++i){
      int u = G[v][i];
      int dd = d + w[v][i];
      if(d1[u] > dd){
        swap(dd, d1[u]);
        pq.push(P(d1[u], u));
      }
      if(d1[u] == dd)  continue;
      if(d2[u] > dd){
        d2[u] = dd;  pq.push(P(d2[u], u));
      }
    }
  }
  return d2[n];
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 1; i <= n; ++i)  G[i].clear(), w[i].clear();
    for(int i = 0; i < m; ++i){
      int u, v, val;
      scanf("%d %d %d", &u, &v, &val);
      G[u].push_back(v);
      G[v].push_back(u);
      w[u].push_back(val);
      w[v].push_back(val);
    }
    printf("%d
", dijkstra());
  }
  return 0;
}