1003. Emergency (25)(Dijkstra + DFS)

这一题算是我对于 Dijkstra 和 DFS 的初步认识吧，学习水准还是比较低的。

贴代码！

#include <cstdio>

#define INF 65535
#define MX 501

int mp[MX][MX];
int v[MX];
int dist[MX];
int teams[MX];
int amount[MX];
int pathcount[MX];
int N,M,start,en;

int main()
{

scanf("%d %d %d %d", &N,&M,&start,&en);
for (int i = 0; i < N; i++)
{
scanf("%d", &teams[i]);
}
for (int i=0; i<N; i++)
{
dist[i] = INF;
for (int j=0; j<N; j++)
mp[i][j] = INF;
}
for (int i=0; i<M; i++)
{
int c1, c2, L;
scanf("%d %d %d", &c1,&c2,&L);
mp[c1][c2] = mp[c2][c1] = L;
}

for (int i = 0; i < N; i++)
{
v[i] = 0;
dist[i] = mp[start][i];

}

pathcount[start] = 1;
dist[start] = 0;
amount[start] = teams[start];

for (int i = 0; i < N; i++)
{
int k = 0,min = INF;
for (int j = 0; j < N; j++)
{
if (!v[j] && dist[j] < min)
{
k = j;
min = dist[j];
}
}
if (min==INF || k==en) break;

v[k] = 1;
for (int j = 0; j < N; j++)
{
if (!v[j])
{
if (min + mp[k][j] < dist[j])
{
dist[j] = min + mp[k][j];
pathcount[j] = pathcount[k];
amount[j] = amount[k] + teams[j];
}
else if (min + mp[k][j] == dist[j])
{
if (amount[j] < amount[k] + teams[j])
amount[j] = amount[k] + teams[j];
pathcount[j] += pathcount[k];
}
}
}
}

printf("%d %d\n", pathcount[en], amount[en]);
return 0;
}

这一题的 Dijkstra 还是有一点的变形，觉得真是一个神奇的算法。

接着就是 DFS 用这种法子觉得很暴力，在 pat 里也发现时间和空间都相比 Dijkstra 比较大。

贴代码

#include <cstdio>

#define MX 501

int N,M,C1,C2;
int teams[MX] = {0};
int mp[MX][MX] = {0};
bool boolDfs[MX] = {false};

int minDistance = 65535;
int shortPath = 0;
int maxTeams = 0;

void dfs(int start, int distance, int team)
{
if (start == C2)
{
if (distance < minDistance)
{
minDistance = distance;
shortPath = 1;
maxTeams = team;
}
else if (distance == minDistance)
{
shortPath++;
if (team > maxTeams)
maxTeams = team;
}
return;
}

boolDfs[start] = true;
for (int i = 0; i < N; i++)
{
if (boolDfs[i] == false && mp[start][i] > 0)
{
dfs(i,distance + mp[start][i],team + teams[i]);
}
}
boolDfs[start] = false;
}

int main()
{
scanf("%d %d %d %d",&N,&M,&C1,&C2);

for (int i = 0; i < N; i++)
{
scanf("%d",&teams[i]);
}
for (int i = 0; i < M; i++)
{
int c1,c2,tmp;
scanf("%d %d %d",&c1,&c2,&tmp);
mp[c1][c2] = mp[c2][c1] = tmp;
}
dfs(C1,0,teams[C1]);

printf("%d %d",shortPath,maxTeams);
return 0;
}

dfs 的思路蛮不错的，可以适用于很多题目，讲道理刷题考pat而言的话，dfs更好一点，pat对于空间要求并不高，可以用dfs 快速解决一些dp 的题目。

学习！

记住！