图论算法 待补充

1. 贝尔福特曼

#include<iostream>
using namespace std;
#include<cstring>
#include<cstdio>
struct edge {
	int s, e, v; //起点，终点，边权
};
edge edg[200005]; //存储两次
int n, m, s, ans[100005], cnt;
void add_edge(int a, int b, int c) {
	edg[cnt].s = a;
	edg[cnt].e = b;
	edg[cnt].v = c;
	cnt++;
}
int main() {
	memset(ans, 0x3f, sizeof(ans));
	scanf_s("%d%d%d", &n, &m, &s);
	for (int i = 0; i < m; i++) {
		int a, b, c;
		scanf_s("%d%d%d", &a, &b, &c);
		add_edge(a, b, c);
		add_edge(b, a, c);
	}
	ans[s] = 0;
	for (int i = 0; i < n; i++) {
		int f = 0;
		for (int j = 0; j < cnt; j++) {
			int e = edg[j].e, s = edg[j].s, v = edg[j].v;
				if (ans[e] > ans[s] + v) {
				ans[e] = ans[s] + v;
				f = 1;
			}
		}
		if (!f)
			break;
	}
	for (int i = 1; i <= n; i++) {
		if (ans[i] == 0x3f3f3f3f)
			puts("-1");
		else
			printf("%d\n", ans[i]);
	}
	return 0;
}

2. 链式前向星+迪杰斯特拉

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
struct edge {
	int e, v, nnext; //终点，权重，下一个点的下标	
};
struct node {
	int now, dis;
	bool operator<(const node& b)const {
		return this->dis > b.dis;
	}
};
edge edg[1000005];
int n, m, s, ans[100005], head[100005], cnt;
void add_edge(int a, int b, int c) {
	edg[cnt].e = b;
	edg[cnt].v = c;
	edg[cnt].nnext = head[a];
	head[a] = cnt++;
}
int main() {
	memset(head, -1, sizeof(head));
	memset(ans, 0x3f, sizeof(ans));
	scanf_s("%d%d%d", &n, &m, &s);
	for (int i = 0; i < m; i++) {
		int a, b, c;
		scanf_s("%d%d%d", &a, &b, &c);
		add_edge(a, b, c);
		add_edge(b, a, c);
	}
	priority_queue<node>que;
	que.push(node{ s,0 });
	ans[s] = 0;
	while (!que.empty()) {
		node temp = que.top();
		que.pop();
		if (ans[temp.now] < temp.dis) {
			continue;
		}
		for (int i = head[temp.now]; i != -1; i = edg[i].nnext) {
			int e = edg[i].e, v = edg[i].v;
			if (ans[e] > ans[temp.now] + v) {
				ans[e] = ans[temp.now] + v;
				que.push(node{ e,ans[e] });
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		if (ans[i] == 0x3f3f3f3f)
			puts("-1");
		else
			printf("%d\n", ans[i]);
	}






	return 0;
}

3. 链式前向星+基于队列优化的贝尔福特曼

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct edge {
	int e, v, nnext;
};
edge edg[200005]; //存储两次
int n, m, s, ans[100005], head[100005], mark[100005], cnt;

void add_edge(int a, int b, int c) {
	edg[cnt].e = b;
	edg[cnt].v = c;
	edg[cnt].nnext = head[a];
	head[a] = cnt++;
}
int main() {
	memset(ans, 0x3f, sizeof(ans));
	memset(head, -1, sizeof(head));
	scanf_s("%d%d%d", &n, &m, &s);
	for (int i = 1; i <= m; i++) {
		int a, b, c;
		scanf_s("%d%d%d", &a, &b, &c);
		add_edge(a, b, c);
		add_edge(b, a, c);
	}
	queue<int> que;
	ans[s] = 0;
	que.push(s);
	mark[s] = 1;
	while (!que.empty()) {
		int temp = que.front();
		que.pop();
		mark[temp] = 0;
		for (int i = head[temp]; i != -1; i = edg[i].nnext) {
			int e = edg[i].e, v = edg[i].v;
			if (ans[e] > ans[temp] + v) {
				ans[e] = ans[temp] + v;
				if (mark[e] == 0) {
					que.push(e);
					mark[e] = 1;
				}
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		if (ans[i] == 0x3f3f3f3f)
			puts("-1");
		else
			printf("%d\n", ans[i]);
	}
	return 0;
}

4. 邻接表+迪杰斯特拉

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
int n, m, s, ans[100005];
struct node {
	int now, dis;
	//小顶堆需要重载大于号
	bool operator<(const node& b)const {
		return this->dis > b.dis;
	}
};
struct edge {
	int e, v; //e终点，v权重
};
int main() {
	memset(ans, 0x3f, sizeof(ans));
	scanf_s("%d%d%d", &n, &m, &s);
	vector<vector<edge> >edg(n + 5, vector<edge>());
	for (int i = 0; i < m; i++) {
		int a, b, c;
		scanf_s("%d%d%d", &a, &b, &c);
		edg[a].push_back(edge{ b,c });
		edg[b].push_back(edge{ a,c });
	}
	priority_queue<node>que;
	que.push(node{ s,0 });
	ans[s] = 0; 
	while (!que.empty()) {
		node temp = que.top();
		que.pop();
		if (ans[temp.now] < temp.dis) {
			continue;
		}
		for (int i = 0; i < edg[temp.now].size(); i++) {
			int e = edg[temp.now][i].e, v = edg[temp.now][i].v;
			if (ans[e] > temp.dis + v) {
				ans[e] = temp.dis + v;
				que.push(node{ e,ans[e] });
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		if (ans[i] == 0x3f3f3f3f)
			puts("-1");
		else
			printf("%d\n", ans[i]);
	}
	return 0;
}

5. 邻接矩阵+弗洛伊德

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int ans[1005][1005], n, m, s;
int main() {
	memset(ans, 0x3F, sizeof(ans));
	scanf_s("%d%d%d", &n, &m, &s);
	for (int i = 1; i <= m; i++) {
		int a, b, c;
		scanf_s("%d%d%d", &a, &b, &c);
		if (ans[a][b] > c) {
			ans[a][b] = c;
			ans[b][a] = c;
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				ans[j][k] = min(ans[j][k], ans[j][i] + ans[i][k]);
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		ans[i][i] = 0;
		if (ans[s][i] == 0x3F3F3F3F) {
			puts("-1");
		}
		else {
			printf("%d\n", ans[s][i]);
		}
	}
	return 0;
}