Tyvj 1221 微子危机——战略

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector> 
#include<queue>
#include<algorithm>
#define mx 1001

using namespace std;
struct orz
{
    int d,p;
    friend bool operator <(orz a,orz b) {return a.d>b.d;}//堆和set里面都只有小于号，所以要用小根堆的话要将<重定向为>
};
struct Edge{
    int to;
    int w;
};
priority_queue < orz > ss;
int flag = 0,v[mx],d[mx],n,m,l;
vector<Edge> edge[mx];
void input(){
    cin>>n>>m>>l;
    int u,v,wei;
    Edge test;
    for(int i = 1;i <= l;i++){
        scanf("%d%d%d",&u,&v,&wei);
        test.to = v;
        test.w = wei;
        edge[u].push_back(test);
        test.to = u;
        edge[v].push_back(test);
    }
    for(int i = 0;i < mx;i++) d[i] = 1000000000;
}
void dij(int s)
{
    
    d[s]=0;
    orz tmp;
    tmp.d=0,tmp.p=s;
    ss.push(tmp);
    flag++;
    int x,dd;
    Edge j;
    while (!ss.empty())//不能只做n次，要一直做到堆空
    {
        tmp=ss.top();
        ss.pop();
        x=tmp.p,dd=tmp.d;
        if (v[x]==flag) continue;//这里一定要判断！！！
        v[x]=flag;
        for (int i = 0;i < edge[x].size();i++){
            
            j = edge[x][i];
            if (d[j.to]>dd+j.w)
            {
                d[j.to]=dd+j.w;
                tmp.d=dd+j.w,tmp.p=j.to;
                ss.push(tmp);
            }
        }

    }
}
int main(){
    input();
    dij(m);
    int ans = 0;
    for(int i = 1;i <= n;i++){
        if(i == m) continue;
        if(d[i] >= 1000000000){
            cout<<"Sth wrong."<<endl;
            return 0;
        }
        ans+=d[i];
    } 
    cout<<ans<<" M(s) are needed."<<endl;
    return 0;
}