NEERC 2013 Dwarf Tower (最短路)

Problem D. Dwarf Tower
Input file: dwarf.in
Output file: dwarf.out
Time limit: 2 seconds
Memory limit: 256 megabytes
Little Vasya is playing a new game named “Dwarf Tower”. In this game there are n different items,
which you can put on your dwarf character. Items are numbered from 1 to n. Vasya wants to get the
item with number 1.
There are two ways to obtain an item:
• You can buy an item. The i-th item costs ci money.
• You can craft an item. This game supports only m types of crafting. To craft an item, you give
two particular different items and get another one as a result.
Help Vasya to spend the least amount of money to get the item number 1.
Input
The first line of input contains two integers n and m (1 ≤ n ≤ 10 000; 0 ≤ m ≤ 100 000) — the number
of different items and the number of crafting types.
The second line contains n integers ci — values of the items (0 ≤ ci ≤ 10
9
).
The following m lines describe crafting types, each line contains three distinct integers ai, xi, yi — ai is
the item that can be crafted from items xi and yi (1 ≤ ai, xi, yi ≤ n; ai ̸ = xi; xi ̸ = yi; yi ̸ = ai).
Output
The output should contain a single integer — the least amount of money to spend.
Examples
dwarf.in dwarf.out
5 3
5 0 1 2 5
5 2 3
4 2 3
1 4 5
2
3 1
2 2 1
1 2 3

2

a，b可以交换c，则a->c,距离c[b]，b->c,距离为c[a]，再加一个超级源点连向每个点，距离为c[i]，然后从源点跑最短路，求出到每个点的最短距离。这个最短距离就是通过交换或不换买到物品i的最小代价。有了这些代价后，枚举得到1的方案， 求最小值即可。

#include<cstdio>#include<map>#include<queue>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<list>#include<set>#include<cmath>using namespace std;const int maxn = 1e4 + 5;const int INF = 1e9;const double eps = 1e-6;typedef unsigned long long ULL;typedef long long LL;typedef pair<LL, LL> P;#define fi first#define se secondLL c[maxn];int n;vector<P> G[maxn];priority_queue<P> q;LL dis[maxn];void dij(){    while(!q.empty())        q.pop();    for(int i = 1;i <= n;i++)        dis[i] = INF;    q.push(P(0, 0));    dis[0] = 0;    while(!q.empty()){        P tem = q.top();        q.pop();        LL total = -tem.fi;        int pos = tem.se;        if(total > dis[pos])            continue;        for(int i = 0;i < G[pos].size();i++){            P edge = G[pos][i];            int to = edge.fi;            LL der = edge.se;            if(total+der < dis[to]){                dis[to] = total + der;                q.push(P(-dis[to], to));            }        }    }}P edge[maxn*10];int main(){    freopen("dwarf.in", "r", stdin);    freopen("dwarf.out", "w", stdout);    int m;    while(scanf("%d%d", &n, &m) != EOF){        for(int i = 0;i <= n;i++) G[i].clear();        for(int i = 1;i <= n;i++){            scanf("%I64d", &c[i]);            G[0].push_back(P(i, c[i]));        }        int cnt = 0;        while(m--){            int to, x, y;            scanf("%d%d%d", &to, &x, &y);            G[x].push_back(P(to, c[y]));            G[y].push_back(P(to, c[x]));            if(to == 1){                edge[cnt++] = P(x, y);            }        }        dij();        LL ans = dis[1];        for(int i = 0;i < cnt;i++){            int x = edge[i].fi;            int y = edge[i].se;            ans = min(ans, dis[x]+dis[y]);        }        printf("%I64d\n", ans);    }    return 0;}