題目鏈接：

http://acm.split.hdu.edu.cn/showproblem.php?pid=3665

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.

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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.

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

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Sample Input 5 1 0 1 1 2 0 2 3 3 1 1 1 4 100 0 1 0 1

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Sample Output 2 Hint: 題意： 一個含有N個小城鎮的大城鎮，以0到N-1的順序，先給出兩個數a,b;表示i到其他城鎮有a條道路,b表示該地方是否有海，0表示沒有海，1表示有海。 然后在給出a1,x;a1表示i到a1之間有道路，距離是x。求從0到有海的城市的最小的距離。 題解： 簡單的最短路問題。 代碼： #include <cmath> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 10+10; #define met(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f int map[maxn][maxn]; int visited[maxn],dis[maxn]; int num[maxn]; int n,m; void dijkstra(int x) {int min=0,p=0;for(int i=0;i<n;i++){dis[i]=map[x][i];visited[i]=0;}visited[x]=1;for(int i=0;i<n;i++){min=inf;for(int j=0;j<n;j++){if(!visited[j]&&dis[j]<min){min=dis[j];p=j;}}visited[p]=1;for(int j=0;j<n;j++){if(!visited[j]&&dis[p]+map[p][j]<dis[j])dis[j]=dis[p]+map[p][j];}} } int main() {while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++)for(int j=0;j<n;j++)map[i][j]=inf;met(num,0);for(int i=0;i<n;i++){scanf("%d%d",&m,&num[i]);for(int j=0;j<m;j++){int x,y;scanf("%d%d",&x,&y);map[i][x]=map[x][i]=y;}}dijkstra(0);int ans=inf;for(int i=0;i<n;i++){if(num[i]==1)ans=min(ans,dis[i]);}printf("%d\n",ans);} }

　　

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轉載于:https://www.cnblogs.com/TAT1122/p/5853273.html

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