最短路三种算法的模板：
1. 弗洛伊德：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long LL;
int G[10][10];
int i, j, k, n, m, x, y;
int main() {
cin >> n >> m;
for(i = 1; i <= n; i++) {
for(j = 1; j <= n; j++) G[i][j] = 1 << 10;
}
for(i = 1; i <= m; i++) {
cin >> x >> y >> k;
G[x][y] = k;
G[y][x] = k;
}
for(i = 1; i <= n; i++) {
for(j = 1; j <= n; j++) {
for(k = 1; k <= n; k++) {
if(i == j || i == k || j == k) continue;
G[j][k] = min(G[j][k], G[j][i] + G[i][k]);
}
}
}
cout << G[1][2] << endl;
return 0;
}
对应题目：POJ3660、POJ3615
POJ - 3660题解：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
int G[107][107];
int n, m, u, v;
int res;
int main() {
while(scanf("%d %d", &n, &m) != EOF) {
memset(G, 0, sizeof(G));
while(m--) {
cin >> u >> v;
G[u][v] = 1;
}
for(int k = 1; k <= n; k++)
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
if(G[i][k] && G[k][j]) G[i][j] = 1;
res = 0;
for(int i = 1; i <= n; i++) {
int j;
for(j = 1; j <= n; j++) {
if(i == j) continue;
if(!G[i][j] && !G[j][i]) break;
}
//cout << tmp << endl;
if(j > n) res++;
}
cout << res << endl;
}
return 0;
}
POJ - 3615题解：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const int Inf = 1 << 30;
int G[330][330];
int n, m, q;
int u, v, w;
int x, y;
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) G[i][j] = Inf;
else G[i][j] = 0;
}
}
while(m--) {
scanf("%d %d %d", &u, &v, &w);
G[u][v] = w;
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
for(int k = 1; k <= n; k++) {
if(G[j][i] < Inf && G[i][k] < Inf) {
int tmp = max(G[j][i], G[i][k]);
G[j][k] = min(tmp, G[j][k]);
}
}
}
}
while(q--) {
scanf("%d %d", &x, &y);
if(G[x][y] == Inf) puts("-1");
else printf("%d\n", G[x][y]);
}
}
2. 迪杰斯特拉：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;
int tu[50][50];
int dis[50], vis[50];
int n, m;
void dij() {
int j, i, u, MAX;
for(i = 2; i <= n; i++) {
dis[i] = tu[1][i];
vis[i] = 0;
}
vis[1] = 1;
for(i = 2; i <= n; i++) {
MAX = 1 << 20;
for(j = 1; j <= n; j++) {
if(vis[j] == 0 && dis[j] < MAX) {
MAX = dis[j];
u = j;
}
}
vis[u] = 1;
for(j = 1; j <= n; j++) {
if(vis[j] == 0 && dis[j] > tu[u][j] + dis[u])
dis[j] = tu[u][j] + dis[u];
}
}
}
int main() {
int i, j, x, y, c;
cin >> n >> m;
for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++) tu[i][j] = 1 << 20;
for(i = 1; i <= m; i++) {
cin >> x >> y >> c;
tu[x][y] = c;
tu[y][x] = c;
}
dij();
cout<< dis[n] << endl;
return 0;
}
对应题目：POJ3268
题解：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int Inf = 1 << 30;
const int maxx = 1e3 + 7;
int G[maxx][maxx];
int dis[maxx], diss[maxx];
bool vis[maxx];
int n, m, p;
int u, v, w;
int Dij() {
int minT;
int u;
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; i++) {
dis[i] = G[p][i];
diss[i] = G[i][p];
}
for(int i = 1; i <= n; i++) {
minT = Inf;
for(int j = 1; j <= n; j++) {
if(!vis[j] && dis[j] < minT) {
minT = dis[j];
u = j;
}
}
vis[u] = 1;
for(int j = 1; j <= n; j++) {
if(!vis[j] && dis[j] > G[u][j] + dis[u])
dis[j] = G[u][j] + dis[u];
}
}
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; i++) {
minT = Inf;
for(int j = 1; j <= n; j++) {
if(!vis[j] && diss[j] < minT) {
minT = diss[j];
u = j;
}
}
vis[u] = 1;
for(int j = 1; j <= n; j++) {
if(!vis[j] && diss[j] > G[j][u] + diss[u])
diss[j] = G[j][u] + diss[u];
}
}
int maxT = -Inf;
for(int i = 1; i <= n; i++) maxT = max(maxT, dis[i] + diss[i]);
return maxT;
}
int main() {
while(scanf("%d %d %d", &n, &m, &p) != EOF) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(i != j) G[i][j] = Inf;
else G[i][j] = 0;
}
}
while(m--) {
cin >> u >> v >> w;
G[u][v] = w;
}
cout << Dij() << endl;
}
return 0;
}
3. SPFA：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MaxN = 1e5;
const LL INF = 1e9;
bool vis[MaxN + 5];
int all;
int n, m;
int pre[2 * MaxN + 5], last[MaxN + 5], other[2 * MaxN + 5], cost[2 * MaxN + 5];
LL dis[MaxN + 5];
void Build(int u, int v, int w) {
pre[++all] = last[u];
last[u] = all;
other[all] = v;
cost[all] = w;
}
void Spfa(int x) {
queue<int>que;
for(int i = 1; i <= n; i++) dis[i] = INF, vis[i] = 0;
vis[x] = 1, dis[x] = 0;
que.push(x);
while(!que.empty()) {
int pos = que.front();
que.pop();
int ed = last[pos];
while(ed != -1) {
int dr = other[ed];
int w = cost[ed];
if(dis[dr] > dis[pos] + w) {
dis[dr] = dis[pos] + w;
if(!vis[dr]) {
vis[dr] = 1;
que.push(dr);
}
}
ed = pre[ed];
}
vis[pos] = 0;
}
}
int main() {
while(~scanf("%d %d", &n, &m)) {
if(n == 0 && m == 0) break;
memset(last, -1, sizeof(last));
all = -1;
for(int i = 1; i <= m; i++) {
int x, y, w;
scanf("%d %d %d", &x, &y, &w);
Build(x, y, w);
Build(y, x, w);
}
Spfa(1);
printf("%d\n", dis[n]);
}
return 0;
}