搜索与图论

树是特殊的图（有向无环图），无向图是特殊有向图，所以讨论有向图即可

有向图存储方式：

• 邻接矩阵：g[a.b]，n*n复杂度适合稠密图，且不能处理重边（一般保留min即可）
• 邻接表：每个点都是一个单链表，存它可以走到哪些点（头插法）

849. Dijkstra求最短路 I

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510;

int n, m;
int g[N][N];
int dist[N];//1号点到其他点的距离
bool st[N];//每个点的最短路是否确定，是否在s集合

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < n; i ++)//遍历n次
    {
        int t = -1;
        for(int j = 1; j <= n; j ++)//找到不在s中d最小的点
            if(!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        st[t] = true;
        for(int j = 1; j <= n; j ++)//遍历所有点更新距离
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main()
{
    scanf("%d%d", &n, &m);
    
    //初始化
    memset(g, 0x3f,sizeof g);
    
    //读入
    while(m --)
    {
        int a,b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = min(g[a][b], c);
    }
    
    int t = dijkstra();
    printf("%d\n", t);
    
    return 0;
}

853. 有边数限制的最短路

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], backup[N];

struct Edge{
    int a, b, w;
}edges[M];

int bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < k; i ++)
    {
        memcpy(backup, dist, sizeof dist);
        for(int j = 0; j < m; j ++)
        {
            int a = edges[j].a, b = edges[j].b, w = edges[j].w;
            dist[b] = min(dist[b], backup[a] + w);
        }
    }
    if(dist[n] > 0x3f3f3f3f / 2 ) return -2;
    return dist[n];
}
int main()
{
    cin>>n>>m>>k;
    for(int i = 0; i < m; i ++)
    {
        int a, b, w;
        cin>>a>>b>>w;
        edges[i] = {a, b, w};
    }
    int t = bellman_ford();
    if(t == -2) puts("impossible");
    else cout<<t<<endl;
    return 0;
}

851. spfa求最短路

854. Floyd求最短路

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

void floyd()
{
    for(int k = 1; k <= n; k ++)
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= n; j ++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
int main()
{
    cin>> n>>m>>Q;
    //初始化距离矩阵
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= n; j ++)
            if(i == j) d[i][j] = 0;
            else d[i][j] = INF;
            
    while(m --)
    {
        int a, b, w;
        cin>>a>>b>>w;
        
        d[a][b] = min(d[a][b], w);
        
    }
    
    floyd();
    
    while(Q--)
    {
        int a, b;
        cin>>a>>b;
        if(d[a][b] > INF / 2) puts("impossible");
        else cout<<d[a][b]<<endl;
    }
    
    return 0;
}