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
#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;
}