备战NOIP——模板复习10
这里只有模板,并不作讲解,仅为路过的各位做一个参考以及用做自己复习的资料,转载注明出处。
最短路算法
Dijkstra + 堆优化
/*Copyright: Copyright (c) 2018
*Created on 2018-11-1
*Author: 十甫
*Version 1.0
*Title: Dijkstra + heap
*Time: 10 mins
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 10005;
const int maxm = 10005;
int head[maxn];
struct edge {
int to, val, next;
} data[maxm * 2];
inline void add_edge(int from, int to, int val, int i) {
data[i] = (edge) {to, val, head[from]};
head[from] = i;
}
int n, m;
int dis[maxn];
bool vis[maxn];
struct node {
int dist, ord;
bool operator < (const node cmp) const {
return dist > cmp.dist;
}
};
void Dij(int p) {
memset(dis, 0x3f, sizeof(dis));
memset(vis, false, sizeof(vis));
dis[p] = 0;
priority_queue <node> q;
q.push((node) {dis[p], p});
for(int i = 1;i < n;i++) {
while(vis[q.top().ord]) q.pop();
node tmp = q.top();
int u = tmp.ord;
q.pop();
vis[u] = true;
for(int i = head[u];i;i = data[i].next) {
int v = data[i].to, k = data[i].val;
if(dis[v] > dis[u] + k) {
dis[v] = dis[u] + k;
q.push((node) {dis[v], v});
}
}
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1;i <= m;i++) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add_edge(a, b, c, i), add_edge(b, a, c, i + m);
}
int p;
scanf("%d", &p);
Dij(p);
for(int i = 1;i <= n;i++) {
printf("%d ", dis[i]);
}
printf("\n");
return 0;
}
Floyd
/*Copyright: Copyright (c) 2018
*Created on 2018-11-1
*Author: 十甫
*Version 1.0
*Title: Floyd
*Time: 4 mins
*/
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int size = 605;
inline int Min(int a, int b) {
return a < b ? a : b;
}
int n, m, dis[size][size];
void Floyd() {
for(int k = 1;k <= n;k++) {
for(int i = 1;i <= n;i++) {
for(int j = 1;j <= n;j++) {
dis[i][j] = Min(dis[i][j], dis[i][k] + dis[k][j]);
}
}
}
}
int main() {
scanf("%d%d", &n, &m);
memset(dis, 0x3f, sizeof(dis));
for(int i = 1;i <= n;i++) {
dis[i][i] = 0;
}
for(int i = 1;i <= m;i++) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
dis[a][b] = dis[b][a] = Min(dis[a][b], c);
}
Floyd();
int q;
scanf("%d", &q);
while(q--) {
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", dis[a][b]);
}
return 0;
}
SPFA
/*Copyright: Copyright (c) 2018
*Created on 2018-11-1
*Author: 十甫
*Version 1.0
*Title: SPFA
*Time: 7 mins
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 10005;
const int maxm = 10005;
int head[maxn];
struct edge {
int to, val, next;
} data[maxm * 2];
inline void add_edge(int from, int to, int val, int i) {
data[i] = (edge) {to, val, head[from]};
head[from] = i;
}
int n, m, dis[maxn];
bool vis[maxn];
void SPFA(int p) {
memset(dis, 0x3f, sizeof(dis));
memset(vis, false, sizeof(vis));
dis[p] = 0, vis[p] = true;
queue <int> q;
q.push(p);
while(!q.empty()) {
int u = q.front();
q.pop();
for(int i = head[u];i;i = data[i].next) {
int v = data[i].to, k = data[i].val;
if(dis[v] > dis[u] + k) {
dis[v] = dis[u] + k;
if(!vis[v]) q.push(v);
}
}
vis[u] = false;
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1;i <= m;i++) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add_edge(a, b, c, i), add_edge(b, a, c, i + m);
}
int p;
scanf("%d", &p);
SPFA(p);
for(int i = 1;i <= n;i++) {
printf("%d ", dis[i]);
}
printf("\n");
return 0;
}
如果图包含负环的话,可以利用SPFA判断负环, 具体做法参考@forever_dreams 的Blog
用 cnt[ i ] 表示从起点(假设就是 1)到 i 的最短距离包含点的个数,初始化 cnt[ 1 ] = 1,那么当我们能够用点 u 松弛点 v 时,松弛时同时更新 cnt[ v ] = cnt[ u ] + 1,若发现此时 cnt[ v ] > n,那么就存在负环
原文链接: SPFA判负环
NOIP 2018 RP++