POJ3114 Country in Wars Tarjan+Dij
2014-08-29 11:17
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题目大意:在一个连通分量的两点间距离为零,问题目给定条件的两点间距离。
题目分析:Tarjan后重建图 用gro_id表示在新图中的新位置,map1[][]表示新图中连通分量之间的距离,输出之即可。
如下代码:
#include<stdio.h>
#include<string.h>
#include<stack>
#include<vector>
#define inf 99999999
using namespace std;
vector<int> vec[505];
stack<int> sta;
int dfn[505],low[505];
int vis[505];
int gro_id[505],gro[505];
int map[505][505],map1[505][505];
int d[505],v[505];
int now,id;
int n,m;
void tarjan(int s)
{
vis[s]=2;
now++;
low[s]=dfn[s]=now;
sta.push(s);
for(int i=0; i<vec[s].size(); i++)
{
if(vis[vec[s][i]]==0)
{
tarjan(vec[s][i]);
low[s]=low[s]<low[vec[s][i]]?low[s]:low[vec[s][i]];
}
else if(vis[vec[s][i]]==2)
low[s]=low[s]<dfn[vec[s][i]]?low[s]:dfn[vec[s][i]];
}
if(low[s]==dfn[s])
{
id++;
while(1)
{
int t=sta.top();
gro_id[t]=id;
vis[t]=1;
sta.pop();
gro[id]++;
if(s==t)
break;
}
}
}
void dij(int start,int end)
{
int min,min_f;
for(int i=1; i<=id; i++)
{
d[i]=map1[start][i];
v[i]=0;
}
v[start]=1;
for(int i=1; i<id; i++)
{
min=inf;
min_f=1;
for(int j=1; j<=id; j++)
{
if(!v[j]&&min>d[j])
{
min=d[j];
min_f=j;
}
}
v[min_f]=1;
for(int j=1; j<=id; j++)
{
if(!v[j]&&d[j]>d[min_f]+map1[min_f][j])
d[j]=d[min_f]+map1[min_f][j];
}
}
if(d[end]==inf)
printf("Nao e possivel entregar a carta\n");
else
printf("%d\n",d[end]);
}
int main()
{
while(scanf("%d%d",&n,&m)&&n!=0)
{
now=id=0;
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(gro_id,0,sizeof(gro_id));
memset(gro,0,sizeof(gro));
for(int i=1; i<=n; i++)
{
vec[i].clear();
map[i][i]=map1[i][i]=0;
for(int j=1; j<=n; j++)
map[i][j]=map[j][i]=map1[i][j]=map1[j][i]=inf;
}
for(int i=0; i<m; i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
vec[a].push_back(b);
if(map[a][b]>c)
map[a][b]=c;
}
for(int i=1; i<=n; i++)
if(!vis[i])
tarjan(i);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(i!=j&&gro_id[i]!=gro_id[j]&&map[i][j]!=inf)
map1[gro_id[i]][gro_id[j]]=map[i][j]<map1[gro_id[i]][gro_id[j]]?map[i][j]:map1[gro_id[i]][gro_id[j]];
scanf("%d",&m);
while(m--)
{
//printf("while %d\n",m);
int a,b;
scanf("%d%d",&a,&b);
if(gro_id[a]==gro_id[b])
printf("0\n");
else
dij(gro_id[a],gro_id[b]);
}
printf("\n");
}
return 0;
}
题目分析:Tarjan后重建图 用gro_id表示在新图中的新位置,map1[][]表示新图中连通分量之间的距离,输出之即可。
如下代码:
#include<stdio.h>
#include<string.h>
#include<stack>
#include<vector>
#define inf 99999999
using namespace std;
vector<int> vec[505];
stack<int> sta;
int dfn[505],low[505];
int vis[505];
int gro_id[505],gro[505];
int map[505][505],map1[505][505];
int d[505],v[505];
int now,id;
int n,m;
void tarjan(int s)
{
vis[s]=2;
now++;
low[s]=dfn[s]=now;
sta.push(s);
for(int i=0; i<vec[s].size(); i++)
{
if(vis[vec[s][i]]==0)
{
tarjan(vec[s][i]);
low[s]=low[s]<low[vec[s][i]]?low[s]:low[vec[s][i]];
}
else if(vis[vec[s][i]]==2)
low[s]=low[s]<dfn[vec[s][i]]?low[s]:dfn[vec[s][i]];
}
if(low[s]==dfn[s])
{
id++;
while(1)
{
int t=sta.top();
gro_id[t]=id;
vis[t]=1;
sta.pop();
gro[id]++;
if(s==t)
break;
}
}
}
void dij(int start,int end)
{
int min,min_f;
for(int i=1; i<=id; i++)
{
d[i]=map1[start][i];
v[i]=0;
}
v[start]=1;
for(int i=1; i<id; i++)
{
min=inf;
min_f=1;
for(int j=1; j<=id; j++)
{
if(!v[j]&&min>d[j])
{
min=d[j];
min_f=j;
}
}
v[min_f]=1;
for(int j=1; j<=id; j++)
{
if(!v[j]&&d[j]>d[min_f]+map1[min_f][j])
d[j]=d[min_f]+map1[min_f][j];
}
}
if(d[end]==inf)
printf("Nao e possivel entregar a carta\n");
else
printf("%d\n",d[end]);
}
int main()
{
while(scanf("%d%d",&n,&m)&&n!=0)
{
now=id=0;
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(gro_id,0,sizeof(gro_id));
memset(gro,0,sizeof(gro));
for(int i=1; i<=n; i++)
{
vec[i].clear();
map[i][i]=map1[i][i]=0;
for(int j=1; j<=n; j++)
map[i][j]=map[j][i]=map1[i][j]=map1[j][i]=inf;
}
for(int i=0; i<m; i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
vec[a].push_back(b);
if(map[a][b]>c)
map[a][b]=c;
}
for(int i=1; i<=n; i++)
if(!vis[i])
tarjan(i);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(i!=j&&gro_id[i]!=gro_id[j]&&map[i][j]!=inf)
map1[gro_id[i]][gro_id[j]]=map[i][j]<map1[gro_id[i]][gro_id[j]]?map[i][j]:map1[gro_id[i]][gro_id[j]];
scanf("%d",&m);
while(m--)
{
//printf("while %d\n",m);
int a,b;
scanf("%d%d",&a,&b);
if(gro_id[a]==gro_id[b])
printf("0\n");
else
dij(gro_id[a],gro_id[b]);
}
printf("\n");
}
return 0;
}
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