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 501int 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 501int 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！