标签：图论 edg temp 补充 d% int 算法 ans include
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;
}标签：图论,edg,temp,补充,d%,int,算法,ans,include 来源： https://blog.csdn.net/weixin_54664477/article/details/122798781
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