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("
");
	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
", 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("
");
	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

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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

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