POJ2139 Six Degrees of Cowvin Bacon 最短路
2014-07-15 20:54
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这题水题,就是将出现在同一电影的牛两两连上一个权为1的边,然后询问任意两头牛之间的“距离”。这里直接用的是floyd求最短路。
#ifndef HEAD #include <stdio.h> #include <vector> #include <math.h> #include <string.h> #include <string> #include <iostream> #include <queue> #include <list> #include <algorithm> #include <stack> #include <map> using namespace std; #endif // !HEAD #ifndef QUADMEMSET inline void QuadMemSet(void* dst, int iSize, int value) { iSize = iSize / 4; int* newDst = (int*)dst; #ifdef WIN32 __asm { mov edi, dst mov ecx, iSize mov eax, value rep stosd } #else for (int i = 0; i < iSize; i++) { newDst[i] = value; } #endif } #endif int Dis[301][301]; void Flyod(int N) { 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() { #ifdef _DEBUG freopen("d:\\in.txt", "r", stdin); #endif int N, M; vector<int> tmp; scanf("%d %d\n", &N, &M); QuadMemSet(Dis, sizeof(Dis), 10000000); for (int i = 0; i < M;i++) { int c; tmp.clear(); scanf("%d", &c); tmp.resize(c); for (int j = 0; j < c;j++) { scanf("%d", &tmp[j]); } for (int j = 0; j < c;j++) { for (int h = 0; h < c;h++) { if (tmp[j] == tmp[h]) { Dis[tmp[j]][tmp[j]] = 0; } else { Dis[tmp[h]][tmp[j]] = 1; Dis[tmp[j]][tmp[h]] = 1; } } } } Flyod(N); int imin = 1000000; for (int i = 1; i <= N;i++) { int sum = 0; for (int j = 1; j <= N;j++) { if (i == j) { continue; } sum += Dis[i][j]; } if (imin > sum) { imin = sum; } } printf("%d\n", imin * 100 / (N - 1)); return 1; }
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