ZOJ 3946 Highway Project (最短路)

题意：单源最短路，给你一些路，给你这些路的长度，给你修这些路的话费，求最短路和最小花费。

析：本质就是一个最短路，不过要维护两个值罢了，在维护花费时要维护的是该路要花多少，而不是总的路线花费。

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 LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 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;
}
struct Node{
  int v, c, d;
};

vector<Node> G[maxn];
struct node{
  int u;  LL d, c;
  node(){ }
  node(int uu, LL dd, LL cc) : u(uu), d(dd), c(cc) { }
  bool operator < (const node &p) const{
    return d > p.d || (d == p.d && c > p.c);
  }
};

LL d[maxn], c[maxn];

void solve(){
  priority_queue<node> pq;
  pq.push(node(0, 0, 0));
  d[0] = c[0] = 0;

  while(!pq.empty()){
    node U = pq.top();  pq.pop();
    int u = U.u;
    for(int i = 0; i < G[u].size(); ++i){
      Node &V = G[u][i];
      int v = V.v;
      if(d[v] > d[u] + V.d){
        d[v] = d[u] + V.d;
        c[v] = V.c;
        pq.push(node(v, d[v], c[v]));
      }
      else if(d[v] == d[u] + V.d && c[v] > V.c){
        c[v] = V.c;
        pq.push(node(v, d[v], c[v]));
      }
    }
  }
  LL ans1 = 0, ans2 = 0;
  for(int i = 1; i < n; ++i){
    ans1 += d[i];
    ans2 += c[i];
  }

  printf("%lld %lld
", ans1, ans2);
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; ++i){
      G[i].clear();
      c[i] = d[i] = LNF;
    }
    for(int i = 0; i < m; ++i){
      int x;
      Node u;
      scanf("%d %d %d %d", &x, &u.v, &u.d, &u.c);
      G[x].push_back(u);
      swap(x, u.v);
      G[x].push_back(u);
    }
    solve();
  }
  return 0;
}