BZOJ 1491: [NOI2007]社交网络( floyd )

floyd...求最短路时顺便求出路径数. 时间复杂度O(N^3) 

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#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

typedef long long ll;

const int maxn = 109;

const int INF = 0X3F3F3F3F;

int N, d[maxn][maxn];

ll cnt[maxn][maxn];

int main() {

int m;

scanf("%d%d", &N, &m);

for(int i = 0; i < N; i++) {

for(int j = 0; j < N; j++)

d[i][j] = INF, cnt[i][j] = 0;

d[i][i] = 1;

cnt[i][i] = 1;

}

while(m--) {

int u, v, w;

scanf("%d%d%d", &u, &v, &w);

u--, v--;

d[u][v] = d[v][u] = w;

cnt[u][v] = cnt[v][u] = 1;

}

for(int k = 0; k < N; k++)

for(int i = 0; i < N; i++)

for(int j = 0; j < N; j++)

if(d[i][k] + d[k][j] < d[i][j]) {

d[i][j] = d[i][k] + d[k][j];

cnt[i][j] = cnt[i][k] * cnt[k][j];

} else if(d[i][k] + d[k][j] == d[i][j])

cnt[i][j] += cnt[i][k] * cnt[k][j];

for(int k = 0; k < N; k++) {

double ans = 0;

for(int i = 0; i < N; i++)

for(int j = 0; j < N; j++) if(d[i][j] != INF)

if(d[i][k] + d[j][k] == d[i][j])

ans += (double) cnt[i][k] * cnt[j][k] / cnt[i][j];

printf("%.3lf\n", ans);

}

return 0;

}

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1491: [NOI2007]社交网络

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 1271  Solved: 727
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Description

Input

Output

输出文件包括n 行，每行一个实数，精确到小数点后3 位。第i 行的实数表 示结点i 在社交网络中的重要程度。

Sample Input

4 4
1 2 1
2 3 1
3 4 1
4 1 1

Sample Output

1.000
1.000
1.000
1.000

HINT

为1

Source