CodeForces 295B - Greg and Graph Floyd的巧妙应用
2013-07-22 15:36
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练习赛时看到这题..我试着用边SPFA边删边的方法做...结果超时..
参考大牛的思路...不断删点找最小路很难有高效的方法.不如把问题反过来..一个空图...每次加点..更新最短路...而Floyd正是枚举中间点更新最短路...所以这个问题用一次Floyd就可以解决了...
Program:
参考大牛的思路...不断删点找最小路很难有高效的方法.不如把问题反过来..一个空图...每次加点..更新最短路...而Floyd正是枚举中间点更新最短路...所以这个问题用一次Floyd就可以解决了...
Program:
#include<iostream> #include<stack> #include<queue> #include<stdio.h> #include<algorithm> #include<string.h> #include<cmath> #define ll long long #define oo 1000000007 #define MAXN 505 using namespace std; int n,del[MAXN]; ll dis[MAXN][MAXN],ans[MAXN]; bool f[MAXN]; int main() { int i,j,x,k; while (~scanf("%d",&n)) { ans[0]=0; for (i=1;i<=n;i++) for (j=1;j<=n;j++) scanf("%I64d",&dis[i][j]),ans[0]+=dis[i][j]; for (i=n;i>=1;i--) scanf("%d",&del[i]); memset(f,false,sizeof(f)); for (x=1;x<=n;x++) { ans[x]=0; k=del[x]; f[k]=true; for (i=1;i<=n;i++) for (j=1;j<=n;j++) { if (dis[i][j]>dis[i][k]+dis[k][j]) dis[i][j]=dis[i][k]+dis[k][j]; if (f[i] && f[j]) ans[x]+=dis[i][j]; } } for (i=n;i>=1;i--) printf("%I64d ",ans[i]); printf("\n"); } return 0; }
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