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题目链接

Sol

这题。。。。我样例没过就A了？？。。算了，就当是样例卡精度吧。。

直接状压dp一下，\(f[sta]\)表示干掉\(sta\)这个集合里面的鸟的最小操作数

转移的时候判断一下一次能干掉多少鸟。。

#include<bits/stdc++.h>

#define LL long long

using namespace std;

const int MAXN = 100001;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, M, a[MAXN], block, siz[MAXN], flag[MAXN], rak[MAXN], tp[MAXN];

vector<int> v[MAXN];

set<int> s[MAXN];

int comp(const int &x, const int &y) {

    return a[x] == a[y] ? x < y : a[x] < a[y];

}

int main() {

    N = read(); M = read(); block = sqrt(M);

    for(int i = 1; i <= N; i++) a[i] = read(), tp[i] = i;

    sort(tp + 1, tp + N + 1, comp);

    for(int i = 1; i <= N; i++) rak[tp[i]] = i;

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

        int x = read(), y = read();

        if(rak[x] > rak[y]) v[x].push_back(y), siz[x]++;

        else v[y].push_back(x), siz[y]++;

    }

    LL ans = 0;

    for(int i = 3; i <= N; i++) {

        int x = tp[i];

        for(int j = 0, to; j < v[x].size(); j++) flag[to = v[x][j]] = i;

        for(int j = 0, to; j < v[x].size(); j++) {

            if(siz[to = v[x][j]] <= block) {

                for(int k = 0; k < v[to].size(); k++)

                    if(flag[v[to][k]] == i) ans += a[x];

            } else {

                for(int k = 0; k < v[x].size(); k++)

                    if(s[to].count(v[x][k])) ans += a[x];

            }

            s[x].insert(to);

        }

    }

    cout << ans;

    return 0;

}

/*

2 1

13 17

2 1

*/