HDU-3790最短路径问题

题目

分析

先按距离求出最短路，再在最短路中找花费最小的路.
引申：多权最短路，在处理好主权的情况下，处理副权。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
#include<algorithm>
#define lson x<<1
#define rson x<<1|1
#define ll long long
#define rint register int
#define mid  ((L + R) >> 1)
using namespace std;
template <typename xxx> inline void read(xxx &x) {
    char c = getchar(),f = 1;x = 0;
    for(;c ^ '-' && !isdigit(c);c = getchar());
    if(c == '-') c = getchar(),f = -1;
    for(;isdigit(c);c = getchar()) x = (x<<1) + (x<<3) + (c ^ '0');
    x *= f;
}
template<typename xxx>void print(xxx x)
{
    if(x<0){putchar('-');x=-x;}
    if(x>9) print(x/10);
    putchar(x%10+'0');
}
const int maxn = 100010;
const int inf = 0x7fffffff;
const int mod = 1e9 + 7;
struct edge{
    int to,last,d,p;
}e[maxn<<1];
int head[maxn],tot;
inline void add(int from,int to,int d,int p) {
    ++tot;
    e[tot].d = d;
    e[tot].p = p;
    e[tot].to = to;
    e[tot].last = head[from];
    head[from] = tot;
}
int n,m;
queue<int>q;
int dis[maxn],vis[maxn],num[maxn];
inline void spfa(int ks) {
    for(rint i = 0;i <= n; ++i) dis[i] = inf,vis[i] = 0,num[i] = 0;
    q.push(ks);dis[ks] = 0;num[ks] = 1;
    while(q.size()) {
        int x = q.front();q.pop();vis[x] = 0;
        for(rint i = head[x]; i; i = e[i].last) {
            if(dis[e[i].to] > dis[x] + e[i].d) {
                dis[e[i].to] = dis[x] + e[i].d;
                if(!vis[e[i].to]) {
                    num[e[i].to] = 1;
                    vis[e[i].to] = 1;
                    q.push(e[i].to);
                }
            }
            else if(dis[e[i].to] == dis[x] + e[i].d) ++num[e[i].to];
        }
    }
}
int cost[maxn];
inline void ddfs(int ks) {
    for(rint i = 0;i <= n; ++i) vis[i] = 0,cost[i] = inf;
    q.push(ks);cost[ks] = 0;
    while(q.size()) {
        int x = q.front();q.pop();vis[x] = 0;
        for(rint i = head[x]; i; i = e[i].last) {
            if(dis[e[i].to] == dis[x] + e[i].d) {
                if(cost[e[i].to] > cost[x] + e[i].p) {
                    cost[e[i].to] = cost[x] + e[i].p;
                    if(!vis[e[i].to]) {
                        vis[e[i].to] = 1;
                        q.push(e[i].to);
                    }
                }
            }
        }
    }
}
int main()
{
    while(1) {
        tot = 0;
        memset(head,0,sizeof(head));
        read(n);read(m);
        if(!n && !m ) break;
        for(rint i = 1;i <= m; ++i) {
            int a,b,c,d;
            read(a);read(b);read(c);read(d);
            if(a == b) continue;
            add(a,b,c,d);add(b,a,c,d);
        }
        int s,t;
        read(s);read(t);
        spfa(s);ddfs(s);
        print(dis[t]),putchar(' '),print(cost[t]),putchar('\n');
    }
    return 0;
}
/*

*/

来源：https://www.cnblogs.com/Thomastine/p/11808652.html