【LOJ】#2077. 「JSOI2016」飞机调度

题解

考虑一架飞机飞完自己之后还能飞到哪些航线，用floyd求两点最短路

这个图建出来是个DAG，求最小路径覆盖即可，二分图匹配

注意判断时是航班的起飞时刻+直飞时间+加油时间+最短路时间

代码

#include <bits/stdc++.h>

#define enter putchar('\n')

#define space putchar(' ')

#define pii pair<int,int>

#define fi first

#define se second

#define mp make_pair

#define MAXN 505

#define mo 99994711

#define pb push_back

#define eps 1e-8

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

        if(c == '-') f = -1;

        c = getchar();

    }

    while(c >= '0' && c <= '9') {

        res = res * 10 + c - '0';

        c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) out(x / 10);

    putchar('0' + x % 10);

}

int N,M;

int g[MAXN][MAXN],P[MAXN],f[MAXN][MAXN];

int X[MAXN],Y[MAXN],D[MAXN];

bool vis[MAXN];

struct node {

    int to,next;

}E[MAXN * MAXN];

int head[MAXN],sumE,matc[MAXN];

void add(int u,int v) {

    E[++sumE].to = v;

    E[sumE].next = head[u];

    head[u] = sumE;

}

bool match(int u) {

    for(int i = head[u] ; i ; i = E[i].next) {

        int v = E[i].to;

        if(!vis[v]) {

            vis[v] = 1;

            if(!matc[v] || match(matc[v])) {

                matc[v] = u;

                return true;

            }

        }

    }

    return false;

}

void Solve() {

    read(N);read(M);

    for(int i = 1 ; i <= N ; ++i) read(P[i]);

    for(int i = 1 ; i <= N ; ++i) {

        for(int j = 1 ; j <= N ; ++j) {

            read(g[i][j]);

        }

    }

    for(int i = 1 ; i <= N ; ++i) {

        for(int j = 1 ; j <= N ; ++j) {

            if(i == j) continue;

            f[i][j] = g[i][j] + P[j];

        }

    }

    for(int k = 1 ; k <= N ; ++k) {

        for(int i = 1 ; i <= N ; ++i) {

            for(int j = 1 ; j <= N ; ++j) {

                f[i][j] = min(f[i][k] + f[k][j],f[i][j]);

            }

        }

    }

    for(int i = 1 ; i <= M ; ++i) {

        read(X[i]);read(Y[i]);read(D[i]);

    }

    for(int i = 1 ; i <= M ; ++i) {

        for(int j = 1 ; j <= M ; ++j) {

            if(i == j) continue;

            if(D[i] + g[X[i]][Y[i]] + P[Y[i]] + f[Y[i]][X[j]] <= D[j]) add(i,j);

        }

    }

    int ans = M;

    for(int i = 1 ; i <= M ; ++i) {

        memset(vis,0,sizeof(vis));

        if(match(i)) --ans;

    }

    out(ans);enter;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}