題意 給出方向（有向）然后判斷從一個點到另一個點的方案數。如果有無數條那么對應位置置為-1

直接先dp處理出來。dp[i][j] = sum(dp[i][k]*dp[k][j]) ?

同時如果兩點之間有無限條路徑。那么這兩點之間必然有一環存在。有f[k][k]!=0?

#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} #define MAXN 35 int dp[MAXN][MAXN]; int N,M; void read() {N = 0;memset(dp,0,sizeof(dp));while (M--){int u,v;scanf("%d%d",&u,&v);dp[u][v] = 1;N = max(N,max(u,v));} } int main() {//freopen("sample.txt","r",stdin);int kase = 0;while (scanf("%d",&M) != EOF){read();for (int k = 0; k <= N; k++)for (int i = 0; i <= N; i++)for (int j = 0 ; j <= N; j++)dp[i][j] += dp[i][k] * dp[k][j];for (int i = 0; i <= N; i++)if (dp[i][i]){for (int j = 0; j <= N; j++)for (int k = 0; k <= N; k++)if (dp[j][i] && dp[i][k]) dp[j][k] = -1;}printf("matrix for city %d\n",kase++);for (int i = 0; i <= N; i++){for (int j = 0; j <= N; j++){if (!j) printf("%d",dp[i][j]);else printf(" %d",dp[i][j]);}putchar('\n');}}return 0; }

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轉載于:https://www.cnblogs.com/Commence/p/4013492.html

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