Codeforces Beta Round #25 (Div. 2 Only)
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题意:
已经给你最短路的邻接矩阵,然后每次给你一条路,连续求出每个点之间的最短路总和。
POINT:
即判断每次能不能进行松弛,因为给你一条路,只能松弛有关这两个点的路。
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define LL long longconst LL maxn = 300+5;LL mp[maxn][maxn];LL ans=0;LL u,v,w;LL n;void doit(){for(LL i=1;i<=n;i++){for(LL j=1;j<=n;j++){if(mp[i][j]>mp[i][u]+mp[u][j]){ans-=mp[i][j]-(mp[i][u]+mp[u][j]);mp[i][j]=mp[j][i]=mp[i][u]+mp[u][j];}if(mp[i][j]>mp[i][v]+mp[v][j]){ans-=mp[i][j]-(mp[i][v]+mp[v][j]);mp[i][j]=mp[j][i]=mp[i][v]+mp[v][j];}}}}int main(){scanf("%lld",&n);for(LL i=1;i<=n;i++){for(LL j=1;j<=n;j++){scanf("%lld",&mp[i][j]);if(i<j) ans+=mp[i][j];}}LL m;scanf("%lld",&m);for(LL i=1;i<=m;i++){scanf("%lld %lld %lld",&u,&v,&w);if(mp[u][v]<=w){printf("%lld ",ans);continue;}else{ans-=mp[u][v]-w;mp[u][v]=mp[v][u]=w;doit();printf("%lld ",ans);}}}
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