7-35 城市间紧急救援 (25 分)

题目链接

Dijkstra：

#include<iostream>
using namespace std;

dfs做法：不知道哪里错了

#include<iostream>
#include<cstring>
using namespace std;
const int N = 510, INF = 0x3f3f3f3f;
int num[N],path[N];
bool book[N],vis[N];//vis剪枝需要
int g[N][N];
int n,m,s,d,minlen = INF,sum,tsum;//sum救援队数量 tsum路径条数
struct bl{
    bool toend;//是否能到终点
    bool renew;//有无更新路径
};
bl dfs(int x,int len,int count,int amount){//当前点,长度,第几条边，救援队数量
	if(vis[x]) return {false,false};
    if(x == d){
        if(len < minlen){
            minlen = len;
            sum = amount;
            tsum = count;
            return {true,true};
        } else if(len == minlen && sum < amount){
            sum = amount;
            tsum = count;
            return {true,true};
        }
        return {true,false};
    }
    int point = -1;
    bool toend = false;
    for(int i = 0; i < n; i++){
        if(!book[i] && g[i][x]!=INF && minlen >= len+g[i][x]){
            book[i] = true;
            bl tmp = dfs(i,len+g[i][x],count+1,amount+num[i]);
            if(tmp.renew) point = i;
            if(tmp.toend) toend = true;
            book[i] = false;
        }
    }
    if(point != -1) {
    	path[count+1] = point;
    	return {true,true};
	} else if(toend == false){ //经过x点无法到达终点就无需再遍历
		vis[x] = true;
		return {false,false};
	} else return{true,false};
}
int main(){
    cin>>n>>m>>s>>d;
    memset(g,0x3f,sizeof(g));
    for(int i = 0; i < n; i++) {
        cin>>num[i];
    }
    for(int i = 0; i < m; i++){
        int u,v,w;
        cin>>u>>v>>w;
        g[u][v] = g[v][u] = w;
    }
    path[0] = s;
    book[s] = true;
    dfs(s,0,0,num[s]);
    cout<<tsum<<' '<<sum<<endl;
    for(int i = 0; i < tsum+1; i++){
        if(i) cout<<' ';
        cout<<path[i];
    }
    return 0;
}

标签：25,false,救援,minlen,int,35,len,return,true
来源： https://blog.csdn.net/hys__handsome/article/details/120636950