【图论】最短路径&&最小生成树问题

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int INF = 0x3fffffff;
//最大顶点数
const int maxn = 1000;
//各个顶点到源点的距离
int dis[maxn];
//邻接矩阵
int g[maxn][maxn];
//顶点和边
int n, m;
//标记是否已经处理过
bool v[maxn];
void dijkstra(int src) {
for(int i = 1; i <= n; i++) {
dis[i] = INF;
}
//源点从1开始
dis[src] = 0;
memset(v, 0, sizeof(v));
for(int i = 1; i <= n; i++) {
int mark = -1, mindis = INF;
for(int j = 1; j <= n; j++) {
if(!v[j] && dis[j] < mindis) {
mindis = dis[j];
mark = j;
}
}
//题目保证存在最短路径
v[mark] = 1;
for(int j = 1; j <= n; j++) {
if(!v[j]) {
dis[j] = min(dis[j], dis[mark]+g[mark][j]);
}
}
}
}

int main()
{
while(scanf("%d%d", &n, &m) != EOF) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
g[i][j] = INF;
}
}
int from, to, d;
for(int i = 1; i <= m; i++) {
scanf("%d%d%d", &from, &to, &d);
g[from][to] = d;
g[to][from] = d;
}
dijkstra(5);
for(int i = 1; i <= n; i++) {
printf("d[%d] = %d\n", i, dis[i]);
}
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 1000;
//顶点数，边数和源点
int n, m, src;
//邻接矩阵
//g[i][j].first表示i节点的第j条边的节点编号
//g[i][j].second 表示边的长度.
vector<pair<int, int> > g[maxn + 10];
//dist[i]表示源点src到i的距离
int dist[maxn + 10];
//是否在队列中
bool inQue[maxn + 10];
queue<int> que;
void spfa() {
memset(inQue, false, sizeof(inQue));
//初始化无穷大
memset(dist, 63, sizeof(dist));
//从src开始
dist[src] = 0;
while(!que.empty()) que.pop();
que.push(src);
inQue[src] = true;
while(!que.empty()) {
int u = que.front();
que.pop();
for(int i = 0; i < (int)g[u].size(); i++) {
if(dist[u] + g[u][i].second < dist[g[u][i].first]) {
dist[g[u][i].first] = dist[u] + g[u][i].second;
if(!inQue[g[u][i].first]) {
inQue[g[u][i].first] = true;
que.push(g[u][i].first);
}
}
}
inQue[u] = false;
}
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF) {
pair<int, int> path;
int from;
for(int i = 1; i <= m; i++) {
scanf("%d%d%d", &from, &path.first, &path.second);
g[from].push_back(path);
swap(from, path.first);
g[from].push_back(path);
}
src = 1;
spfa();
for(int i = 1; i <= n; i++) {
printf("d[%d] = %d\n", i, dist[i]);
}
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
const int INF = 0x3fffffff;
const int N = 1000;
bool v
;
int dist
;
vector<pair<int, int> > g[N + 10];
int n, m;//顶点和边
int Prime() {
//标记是否访问过
memset(v, 0, sizeof(v));
for(int i = 1; i <= N; i++) {
dist[i] = INF;
}
//任找一个顶点，比如1
dist[1] = 0;
int ans = 0;
for(int i = 1; i <= n; ++i) {
int mark = -1;
for(int j = 1; j <= n; ++j) {
if(!v[j]) {
if(mark == -1) mark = j;
else if(dist[j] < dist[mark]) mark = j;
}
}
if(mark == -1) break;
//标记为已经加入的节点
v[mark] = 1;
ans += dist[mark];
for(int j = 0; j < (int)g[mark].size(); ++j) {
if(!v[g[mark][j].first]) {
int x = g[mark][j].first;
dist[x] = min(dist[x], g[mark][j].second);
}
}
}
return ans;
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF) {
for(int i = 1; i <= n; i++) {
g[i].clear();
}
pair<int, int> path;
int from;
for(int i = 1; i <= m; i++) {
scanf("%d%d%d", &from, &path.first, &path.second);
g[from].push_back(path);
swap(from, path.first);
g[from].push_back(path);
}
int ans = Prime();
printf("ans = %d\n", ans);
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1000;
const int maxm = 1000000;
struct edge {
int x, y, w;
edge(int x = 0, int y = 0, int w = 0):x(x), y(y), w(w){}
}e[maxm];
int fa[maxn];
int getfather(int x) {
if(x == fa[x]) return x;
return fa[x] = getfather(fa[x]);
}
//顶点和边
int n, m;
bool cmp(edge a, edge b) {
if(a.w < b.w) return true;
return false;
}
int kruscal() {
int ans = 0;
sort(e, e+m, cmp);
//初始为n个孤立的点
int cnt = n;
for(int i = 1; i <= n; i++) {
fa[i] = i;
}
for(int i = 0; i < m; i++) {
int t1 = getfather(e[i].x);
int t2 = getfather(e[i].y);
if(t1 != t2) {
fa[t1] = t2;
//两个孤立部分合并之后减1
cnt--;
ans += e[i].w;
if(cnt == 1) break;
}
}
return ans;
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF) {
for(int i = 0; i < m; i++) {
scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].w);
}
int ans = kruscal();
printf("ans = %d\n", ans);
}
return 0;
}