【KM匹配】 HDOJ 2448 Mining Station on the Sea

来源：互联网发布：孔浩 java 编辑：程序博客网时间：2024/05/20 00:13

先用floyd预处理出最短路，然后直接km匹配就可以了。。。注意预处理的时候由于只能到达港口一次，所以floyd要先枚举中间点，不能最后枚举中间点。。。

#include <iostream>#include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 305#define maxm 400005#define eps 1e-10#define mod 1000000007#define INF 999999999#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headint G[maxn][maxn];int g[maxn][maxn];int linker[maxn];int slack[maxn];bool visx[maxn];bool visy[maxn];int lx[maxn];int ly[maxn];int a[maxn];int n, m, o, m1, m2, nx, ny;bool dfs(int x){visx[x] = true;for(int y = 0; y < ny; y++) {if(visy[y]) continue;int tmp = lx[x] + ly[y] - G[x][y];if(!tmp) {visy[y] = true;if(linker[y] == -1 || dfs(linker[y])) {linker[y] = x;return true;}}else slack[y] = min(slack[y], tmp);}return false;}int km(){memset(linker, -1, sizeof linker);memset(ly, 0, sizeof ly);for(int i = 0; i < nx; i++) {lx[i] = -INF;for(int j = 0; j < ny; j++)lx[i] = max(lx[i], G[i][j]);}for(int x = 0; x < nx; x++) {for(int i = 0; i < ny; i++)slack[i] = INF;while(true) {memset(visx, 0, sizeof visx);memset(visy, 0, sizeof visy);if(dfs(x)) break;int d = INF;for(int i = 0; i < ny; i++)if(!visy[i] && d > slack[i])d = slack[i];for(int i = 0; i < nx; i++)if(visx[i]) lx[i] -= d;for(int i = 0; i < ny; i++)if(visy[i]) ly[i] += d;else slack[i] -= d;}}int res = 0;for(int i = 0; i < ny; i++) res += G[linker[i]][i];return res;}void read(){scanf("%d%d%d", &m, &m1, &m2);o = n + m;for(int i = 0; i < o; i++)for(int j = 0; j < o; j++)g[i][j] = INF;for(int i = 0; i < n; i++) scanf("%d", &a[i]), a[i]--, a[i] += n;int u, v, w;while(m1--) {scanf("%d%d%d", &u, &v, &w);u--, v--;u += n;v += n;g[u][v] = min(g[u][v], w);g[v][u] = min(g[v][u], w);}while(m2--) {scanf("%d%d%d", &u, &v, &w);u--, v--;v += n;g[u][v] = min(g[u][v], w);g[v][u] = min(g[v][u], w);}}void work(){for(int k = n; k < o; k++)for(int i = 0; i < o; i++)for(int j = 0; j < o; j++)g[i][j] = min(g[i][j], g[i][k] + g[k][j]);for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)if(g[a[i]][j] == INF) G[i][j] = 0;else G[i][j] = -g[a[i]][j];nx = ny = n;printf("%d\n", -km());}int main(){while(scanf("%d", &n)!=EOF) {read();work();}return 0;}

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