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 second

LL 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;
}


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