[牛客][北大考研复试]I Wanna Go Home[dijkstra]

题目描述

    The country is facing a terrible civil war----cities in the country are divided into two parts supporting different leaders. As a merchant, Mr. M does not pay attention to politics but he actually knows the severe situation, and your task is to help him reach home as soon as possible.     "For the sake of safety,", said Mr.M, "your route should contain at most 1 road which connects two cities of different camp."     Would you please tell Mr. M at least how long will it take to reach his sweet home?

输入描述:

    The input contains multiple test cases.
    The first line of each case is an integer N (2<=N<=600), representing the number of cities in the country.
    The second line contains one integer M (0<=M<=10000), which is the number of roads.
    The following M lines are the information of the roads. Each line contains three integers A, B and T, which means the road between city A and city B will cost time T. T is in the range of [1,500].
    Next part contains N integers, which are either 1 or 2. The i-th integer shows the supporting leader of city i. 
    To simplify the problem, we assume that Mr. M starts from city 1 and his target is city 2. City 1 always supports leader 1 while city 2 is at the same side of leader 2. 
    Note that all roads are bidirectional and there is at most 1 road between two cities.
Input is ended with a case of N=0.

输出描述:

    For each test case, output one integer representing the minimum time to reach home.
    If it is impossible to reach home according to Mr. M's demands, output -1 instead.

示例1

输入

复制

2
1
1 2 100
1 2
3
3
1 2 100
1 3 40
2 3 50
1 2 1
5
5
3 1 200
5 3 150
2 5 160
4 3 170
4 2 170
1 2 2 2 1
0

输出

复制

100
90
540

 

单源最短路问题，直接套了裸的Dijkstra（按照惯例加了堆优化），更新dis时加一个2城市不能到1城市的判断即可。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int N = 700;
const int M = 80000;
int n,m,tol;
struct Edge{
    int v,w,next;
}edge[M];
int head[N],vis[N],dis[N];
int lable[N];
void init(){
    tol = 0;
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w){
    edge[tol] = Edge{v,w,head[u]};
    head[u] = tol++;
}
struct Node{
    int u,w;
    bool operator < (const Node a)const{
        return w > a.w;
    }
}; 
priority_queue <Node> q;
void dijkstra(){
    memset(dis,inf,sizeof(dis));
    dis[1] = 0;
    q.push(Node{1,0});
    while(!q.empty()){
        Node a = q.top();
        q.pop();
        int u = a.u,w = a.w;
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = head[u];i != -1;i = edge[i].next){
            int v = edge[i].v,w = edge[i].w;
            if (!vis[v] && dis[v] > dis[u] + w && !(lable[u] == 2 && lable[v] == 1)){
                dis[v] = dis[u] + w;
                q.push(Node{v,dis[v]});
            }
            
        }
    }
}

int main(){
    while (~scanf("%d",&n)&&n){
        scanf("%d",&m);
        init();
        for (int i = 0;i < m;++i){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        for (int i = 1;i <= n;++i){
            int tmp;
            scanf("%d",&tmp);
            lable[i] = tmp;
        }
        dijkstra();
        if (dis[2] == inf) puts("-1");
            else printf("%d\n",dis[2]);
    }
}

 

转载于:https://www.cnblogs.com/mizersy/p/11613946.html