floyd算法入门

floyd算法是一种多源最短路径算法，代码贼短，只有4行，复杂度贼高O(n^3),n是顶点数

#define min(a,b) (a<b?a:b)
const int maxN = 101, inf = 0x3f3f3f3f;
int dist[maxN][maxN], n;
void floyd() {
	for (int k = 1; k <= n; ++k)
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
				dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
}

dist[i][j]表示从i到j的最短路径，我们看到松弛条件dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]),表示从i到j的最短路是经过k短还是直接过去短

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练习

poj1125

题目大意：指定的人之间可以传话，需要一定时间，问谁传给其他的人花的时间最短

解法：模板题

#include<iostream>
#include<cstring>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
const int maxN=101,inf=0x3f3f3f3f;
int dist[maxN][maxN],n;
void floyd(){
	for(int k=1;k<=n;++k)
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
}
int main(){
	int m,a,b;
	while(scanf("%d",&n)&&n){
		memset(dist,inf,sizeof(dist));
		for(int i=1;i<=n;++i){
			scanf("%d",&m);
			while(m--){
				scanf("%d%d",&a,&b);
				dist[i][a]=b;
			}
		}
		floyd();
		int minLength=inf,maxLength,pos;
		for(int i=1;i<=n;++i){
			maxLength=0;
			for(int j=1;j<=n;++j)
				if(i!=j)maxLength=max(maxLength,dist[i][j]);
			if(minLength>maxLength){
				minLength=maxLength;
				pos=i;
			}
		}
		printf("%d %d\n",pos,minLength);
	}
	return 0;
}

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poj3660

题目大意：有n头牛，题目会告诉你谁等级高，谁等级低，问，能确定几头牛的等级

解法：如果某头牛的度为n-1则可以确定，因为假设x+y=n-1，那么这头牛排在x+1或者是y+1这个位置，然后要改松弛条件，如果A>B,B>C,则AC才进行松弛

扫描二维码关注公众号，回复： 1946476 查看本文章

#include<iostream>
#include<cstring>
const int maxN = 101;
bool dist[maxN][maxN], n;
void floyd() {
	for (int k = 1; k <= n; ++k)
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)dist[i][j] = dist[i][j] || dist[i][k] && dist[k][j];
}
int main() {
	int m, a, b, result, ans;
	while (scanf("%d%d", &n, &m) != EOF) {
		while (m--) {
			scanf("%d%d", &a, &b);
			dist[a][b] = 1;
		}
		floyd();
		ans = 0;
		for (int i = 1; i <= n; ++i) {
			result = 0;
			for (int j = 1; j <= n; ++j)if (i != j && (dist[i][j] || dist[j][i]))++result;
			if (result == n - 1)++ans;
		}
		printf("%d\n", ans);
	}
	return 0;
}

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xujcoj1068

题目大意：中文题目

解法：存的之后只存小于的，也就是说要把大于的转换成小于的，然后开一个数组，记录哪些是题目给的，最后遍历输出没访问过的

#include<iostream>
#include<cstring>
const int maxN = 26;
bool visit[maxN][maxN],dist[maxN][maxN];
void floyd() {
	for(int k=0;k<maxN;++k)
		for(int i=0;i<maxN;++i)
			for(int j=0;j<maxN;++j)
				if (!dist[i][j] && dist[i][k] && dist[k][j])dist[i][j] = true;
}
int main() {
	char s[4];
	int t, n;
	scanf("%d", &t);
	for (int i = 1; i <= t; ++i) {
		scanf("%d", &n);
		memset(visit, false, sizeof(visit));
		memset(dist, false , sizeof(dist));
		while (n--) {
			scanf("%s", s);
			visit[s[0] - 'A'][s[2] - 'A'] = visit[s[2] - 'A'][s[0] - 'A'] = true;
			if (s[1] == '<')dist[s[0] - 'A'][s[2] - 'A'] = true;
			else dist[s[2] - 'A'][s[0] - 'A'] = true;
		}
		floyd();
		printf("Case %d:\n", i);
		bool isNone = false;
		for (int i = 0; i < maxN; ++i)
			for (int j = 0; j < maxN; ++j)
				if (dist[i][j] && !visit[i][j]) {
					printf("%c<%c\n", 'A' + i, 'A' + j);
					isNone = true;
				}
		if (!isNone)printf("NONE\n");
	}
	return 0;
}