所谓的日常 #5

大家期末加油~不挂科后面才能好好的做题哇=w=

div.2

CodeForces 330A Cakeminator

给一个r(<=10)行c(<=10)列的矩形蛋糕，每次你会选择一行或一列全部吃掉。有些位置有草莓，然而你不想吃草莓，问最多能吃到多少蛋糕。

如果我们吃了a行b列的话，可以这么算：先数a * m个，再数b * n个，然后把重复算的部分去掉，也就是a * b个。

或者另外开一个标记数组，把吃掉的位置标记一下也是可以的。怎么写高兴怎么来。

#include <stdio.h>

const int N = 10 + 5;
int n,m;
char str[N][N];
bool A[N],B[N];

int work() {
    for (int i = 0; i < n; ++ i) {
        for (int j = 0; j < m; ++ j) {
            if (str[i][j] == 'S') {
                A[i] = true;
                B[j] = true;
            }
        }
    }
    int a = 0,b = 0;
    for (int i = 0; i < n; ++ i)
        if (A[i] == false)
            a ++;
    for (int i = 0; i < m; ++ i) 
        if (B[i] == 0)
            b ++;

    return a * m + b * n - a * b;
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i = 0; i < n; ++ i)
        scanf("%s",str[i]);
    printf("%d
",work());
}

div.1

CodeForces 543B Destroying Roads

给一张n(<=3000)个点的无向连通图，边权为1。问删掉最多的边，使得从点s1到点t1距离不超过l1，从点s2到点t2距离不超过l2。

容易知道如果只有一组限制条件的话(s1,t1,l1)，那么只要把s1到t1的最短路必须经过的边保留下来就可以了。

对于两组限制条件，s1到t1的路径会与s2到t2的路径有重复一部分。那么我们就枚举重复部分的起点和终点，就可以按上面只有一组条件的情况来做了。O(n^2)。

#include <bits/stdc++.h>
using LL = long long ;

const int N = 3000 + 5;
const int INF = 0x3f3f3f3f;

std::vector<int> edges[N];

int n,m;
int points[4],l[2];
int dis[N][N];

int bfs(int source,int *dis) {
    std::queue<int> que;
    std::fill(dis,dis + n,INF);
    dis[source] = 0;
    que.push(source);
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        for (int v : edges[u]) {
            if (dis[v] == INF) {
                dis[v] = dis[u] + 1;
                que.push(v);
            }
        }
    }
}

int work() {
    for (int i = 0; i < n; ++ i) {
        bfs(i,dis[i]);
    }
    if (dis[points[0]][points[1]] > l[0] 
            || dis[points[2]][points[3]] > l[1]) {
        return -1;
    }
    int answer = dis[points[0]][points[1]]
        + dis[points[2]][points[3]];
    for (int i = 0; i < n; ++ i) {
        for (int j = 0; j < n; ++ j) {
            if (dis[points[0]][i] + dis[i][j] + dis[j][points[1]]
                    <= l[0]
                    &&
                    dis[points[2]][i] + dis[i][j] +
                    dis[j][points[3]]
                    <= l[1]) {
                answer = std::min(answer,
                        dis[points[0]][i]
                        + dis[i][j]
                        + dis[j][points[1]]
                        + dis[points[2]][i]
                        + dis[j][points[3]]);
            }
            if (dis[points[0]][i] + dis[i][j] + dis[j][points[1]]
                    <= l[0]
                    && 
                    dis[points[2]][j] + dis[i][j]
                    + dis[i][points[3]]
                    <= l[1]) {
                answer = std::min(answer,
                        dis[points[0]][i]
                        + dis[i][j]
                        + dis[j][points[1]]
                        + dis[points[2]][j]
                        + dis[i][points[3]]);
            }
        }
    }
    return m - answer;
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i = 0; i < m; ++ i) {
        int a,b;
        scanf("%d%d",&a,&b); a --; b --;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }
    scanf("%d%d%d%d%d%d",points + 0,points + 1,l + 0,
            points + 2,points + 3,l + 1);
    for (int i = 0; i < 4; ++ i) {
        points[i] --;
    }
    printf("%d
",work());
}