今日语录：每一次挑战都是一次成长的机会
文章目录
- 朴素DIjkstra
- 堆优化的Dijkstra
- Ballman-Ford
- Floyd
- Spfa(求最短路)
- Spfa（求是否含有负权）
如上所示即为做题时应对的方法
朴素DIjkstra
引用与稠密图，即m<n^2
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 510;
int n, m;
int g[N][N];
int dist[N];
bool st[N];
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
// 将距离初始化为无穷大
dist[1] = 0;
for (int i = 0; i < n; i++)
{
int t = -1;
for (int j = 1; j <= n; j++)
if (!st[j] && (t == -1 || dist[t] > dist[j]))
t = j;
st[t] = true;
for (int j = 1; j <= n; j++)
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);
memset(g, 0x3f, sizeof g);
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;
}
堆优化的Dijkstra
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef pair<int, int>PII;
const int N = 10010;
int n, m;
int h[N],w[N],e[N],ne[N],idx;
int dist[N];
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a],h[a]=idx++;
}
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
//初始化距离为无穷
dist[1] = 0;
priority_queue<PII, vector<PII>, greater<PII>>heap;
heap.push({ 0,1 });
while (heap.size())
{
auto t = heap.top();
heap.pop();
int ver = t.second, distance = t.first;
if (st[ver])continue;
for (int i = h[ver]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > distance + w[i])
{
dist[j] = distance + w[i];
heap.push({ dist[j],j });
}
}
}
if (dist[n] == 0x3f3f3f3f)return -1;
return dist[n];
}
int main()
{
scanf(" % d % d", &n, &m);
memset(h, -1, sizeof h);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
}
int t = dijkstra();
printf("%d\n", t);
return 0;
}
Ballman-Ford
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N = 510,M = 10010;
int m, n,k;
int dist[N], backup[N];
struct Edge
{
int a, b, w;
}edges[M];
int bellman_ford()
{
memset(dist, 0x3f, sizeof dist);
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 -1;
return dist[n];
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < m; i++)
{
int a, b, w;
scanf("%d%d%d", &a, &b, &w);
edges[i] = { a,b,w };
}
int t = bellman_ford();
if (t == -1)puts("impossible");
else printf("%d\n", t);
return 0;
}
Floyd
#include<iostream>
#include<cstring>
#include<algorithm>
#define _CRT_SECURE_NO_WARNINGS
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()
{
scanf("%d%d%d", &n, &m, &Q);
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;
scanf("%d%d%d", &a, &b, &w);
d[a][b] = min(d[a][b], w);
}
floyd();
while (Q--)
{
int a, b;
scanf("%d%d", &a, &b);
if (d[a][b] > INF / 2)puts("impossible");
else printf("%d\n", d[a][b]);
}
return 0;
}
Spfa(求最短路)
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef pair<int, int>PII;
const int N = 10010;
int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
int spfa()
{
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
queue<int>q;
q.push(1);
st[1] = true;
while (q.size())
{
int t = q.front();
q.pop();
st[t] = false;
for (int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
if (!st[j])
{
q.push(j);
st[j] = true;
}
}
}
}
if (dist[n] == 0x3f3f3f3f)return -1;
return dist[n];
}
int main()
{
scanf(" % d % d", &n, &m);
memset(h, -1, sizeof h);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
}
int t = spfa();
printf("%d\n", t);
return 0;
}
Spfa（求是否含有负权）
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#define _CRT_SECURE_NO_WARNINGS
using namespace std;
typedef pair<int, int>PII;
const int N = 10010;
int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N],cnt[N];
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
int spfa()
{
queue<int>q;
for (int i = 1; i <= n; i++)
{
st[i] = true;
q.push(i);
}
while (q.size())
{
int t = q.front();
q.pop();
st[t] = false;
for (int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
cnt[j] = cnt[t] + 1;
if (cnt[j] >= n)return true;
if (!st[j])
{
q.push(j);
st[j] = true;
}
}
}
}
return false;
}
int main()
{
scanf(" % d % d", &n, &m);
memset(h, -1, sizeof h);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
}
if (spfa())puts("Yes");
else puts("No");
return 0;
}
