HDU 1879 繼續暢通工程

來源:互聯網
上載者:User

 

http://acm.hdu.edu.cn/showproblem.php?pid=1879

 

解題思路:這道題給出的資料裡有一些路是已經建好的,所以這條路的成本就不計入總成本中,所以把它直接設定為0。這道題要注意的就是這個了,然後又是prim演算法,我暈哦,暢通工程怎麼都這個演算法。

 

#include <stdio.h><br />#include <string.h><br />#define MaxSize 105<br />#define INIT 999999999</p><p>int Graph[MaxSize][MaxSize];<br />long sum;<br />bool visited[MaxSize];</p><p>int NumOfVillage;</p><p>void Prim()<br />{<br />int i,j;<br />sum = 0;<br />int dist[MaxSize];<br />int min,locate;<br />memset(visited,0,sizeof(visited));<br />for(i=1;i<=NumOfVillage;i++)<br />dist[i] = Graph[1][i];<br />visited[1] = true;<br />for (j=1;j<=NumOfVillage;j++)<br />{<br />min = INIT;<br />for (i=1;i<=NumOfVillage;i++)<br />{<br />if(!visited[i]&&dist[i]<min)<br />{<br />min = dist[i];<br />locate = i;<br />}<br />}<br />visited[locate] = true;<br />for (i=1;i<=NumOfVillage;i++)<br />{<br />if (!visited[i]&&dist[i]>Graph[locate][i])<br />{<br />dist[i] = Graph[locate][i];<br />}<br />}</p><p>}<br />for (i=1;i<=NumOfVillage;i++)<br />{<br />sum+=dist[i];<br />}<br />}</p><p>int main()<br />{<br />int i,j,N,status,weight;<br />while (scanf("%d",&NumOfVillage)!=EOF&&NumOfVillage)<br />{<br />N = (NumOfVillage*(NumOfVillage-1))/2;<br />for (i=1;i<=NumOfVillage;i++)<br />{<br />for (j=1;j<=NumOfVillage;j++)<br />{<br />if(i==j)<br />Graph[i][j] = 0;<br />else<br />Graph[i][j] = INIT;<br />}<br />}<br />while (N--)<br />{<br />scanf("%d%d%d%d",&i,&j,&weight,&status);<br />if(!status)<br />{<br />Graph[i][j] = weight;<br />Graph[j][i] = weight;<br />}<br />else<br />{<br />Graph[i][j] = 0;<br />Graph[j][i] = 0;<br />}<br />}<br />Prim();<br />printf("%ld/n",sum);<br />}<br />return 0;<br />}

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