HDU3665:Seaside

點擊打開題目鏈接

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 955    Accepted Submission(s): 679

Problem Description

XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ’0’. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered ’0’, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.

 

Input

There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers S_Mi and L_Mi, which means that the distance between the i-th town and the S_Mi town is L_Mi.

 

Output

Each case takes one line, print the shortest length that XiaoY reach seaside.

 

Sample Input

5 1 0 1 1 2 0 2 3 3 1 1 1 4 100 0 1 0 1

 

Sample Output

2

 

Source

2010 Asia Regional Harbin

 

Recommend

lcy

 

=====================================題目大意=====================================

已知每個城鎮是否臨海以及從該城鎮出發的所有道路的長度和終點，計算從小Y家（編號爲0的城鎮）去海邊的最短路線長度。

=====================================算法分析=====================================

引入編號爲N的城鎮"海”，爲臨海的城鎮增加長度爲0終點爲“海”的道路，然後Dijkstra求解城鎮0至城鎮N的最短路即可。

=======================================代碼=======================================

#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

const int INF1=0x1f;
const int MAXN=105;

int N,Edge[MAXN][MAXN],Dis[MAXN];

bool Vis[MAXN];

struct NODE  
{  
	NODE(int P,int D) { Pt=P;  Dis=D; }
    friend bool operator < (const NODE& A,const NODE& B) { return A.Dis>B.Dis; }  
	int Pt,Dis;
};  

void Dijkstra()
{
	memset(Vis,0,sizeof(Vis));
	memset(Dis,INF1,sizeof(Dis));
	priority_queue<NODE>q;
	q.push(NODE(0,Dis[0]=0));
    while(!q.empty())  
    {  
        NODE cur=q.top();  q.pop();  
        if(cur.Pt==N)   { return; }  
		if(Vis[cur.Pt]) { continue; }
		Vis[cur.Pt]=1;
        for(int tmp=0;tmp<=N;++tmp)  
        {   
            if(Edge[cur.Pt][tmp]<Dis[tmp]-cur.Dis)     
            {  
                Dis[tmp]=cur.Dis+Edge[cur.Pt][tmp];
                q.push(NODE(tmp,Dis[tmp]));  
            }  
        }  
    }  
} 

int main()
{
	while(scanf("%d",&N)==1)
	{
		memset(Edge,INF1,sizeof(Edge));
		for(int i=0;i<N;++i)
		{
			int M,P;
			scanf("%d%d",&M,&P);
			if(P) { Edge[i][N]=0; }
			while(M--)
			{
				int S,L;
				scanf("%d%d",&S,&L);
				if(Edge[i][S]>L) { Edge[i][S]=L; }
			}
		}
		Dijkstra();
		printf("%d\n",Dis[N]);
	}
	return 0;
}