SNOI2017炸弹

这个东西其实我是不太会的……但是勉强卡过去了。

首先肯定是建有向图，然后求每个节点能访问的节点个数，最裸的打法就是按照题意枚举建边然后tarjan缩点，用bitset记录一下访问节点，但是bitset开不了那么大，只能拿到50%的分，至于开数组枚举我没有试，目测高不了多少。

然后思考这样一个问题，我建的那么多边真的有用么。于是只建离他最近的两个点的边，然后直接topsort统计ans，能拿到80%的数据。

然后就不会了……

去loj看一眼，有一组小的hack数据，异常难受。

看看别人的打法，不是直接统计ans，而是记录这个点所能到达的最左点l[x]和最右点r[x]，然后r[x]-l[x]+1就是他能引爆的炸弹数。可以AC。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <bitset>
#include <map>
#define ll long long
using namespace std;
struct EDGE {
    int ed, nex;
} edge[], edgec[];
int first[], firstc[], numc, num;
struct Point {
    ll x, r;
} p[];
ll read() {
    ll sum = ;
    int f = ;
    char x = getchar();
    while (x < '' || x > '') {
        if (x == '-')
            f = -;
        x = getchar();
    }
    while (x >= '' && x <= '') {
        sum = sum *  + x - '';
        x = getchar();
    }
    return sum * f;
}
ll n, ans1;
const int mod = 1e9 + ;
const int inf=0x7fffffff;
int dfn[], low[], sta[], bl[], du[];
int ord, sccnum, top,l[],r[];
vector<int> scc[];
bool ins[];
void add(int st, int ed) {
    //    cout<<"st="<<st<<" ed="<<ed<<endl;
    edge[++num].ed = ed;
    edge[num].nex = first[st];
    first[st] = num;
}
void addc(int st, int ed) {
    //    cout<<"stc="<<st<<" edc="<<ed<<endl;
    edgec[++numc].ed = ed;
    edgec[numc].nex = firstc[st];
    firstc[st] = numc;
}
void topsort() {
    queue<int> q;
    for (int i = ; i <= sccnum; i++)
        if (!du[i])
            q.push(i);
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        for (int i = firstc[x]; i; i = edgec[i].nex) {
            int y = edgec[i].ed;
            du[y]--;
            l[y]=min(l[y],l[x]);
            r[y]=max(r[y],r[x]);
            if (!du[y])
                q.push(y);
        }
    }
}
void tarjan(int x) {
    dfn[x] = low[x] = ++ord;
    sta[++top] = x;
    ins[x] = ;
    for (int i = first[x]; i; i = edge[i].nex) {
        int y = edge[i].ed;
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        } else if (ins[y])
            low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
        sccnum++;
        int p;
        do {
            p = sta[top--];
            ins[p] = ;
            l[sccnum]=min(l[sccnum],p);
            r[sccnum]=max(r[sccnum],p);
            bl[p] = sccnum;
            scc[sccnum].push_back(p);
        } while (x != p);
    }
}
int main() {
    n = read();
    for (int i = ; i <= n; i++) {
        p[i].x = read();
        p[i].r = read();
    }
    for(int i=;i<=n;i++){
        l[i]=inf;
        r[i]=-inf;
    }
    for (int i = ; i <= n; i++){
        for (int j = i + ; j <= n; j++) {
            if (j == i)
                continue;
            if (p[j].x - p[j].r <= p[i].x ) {
                add(j, i);
                break;
            }
        }
    }
    for (int i = n; i >= ; i--){
        for (int j = i - ; j >= ; j--) {
            if (j == i)
                continue;
            if ( p[i].x <= p[j].x + p[j].r) {
                add(j, i);
                break;
            }
        }
    }
    /*    for(int i=1;i<=n;i++)
                    for(int j=1;j<=n;j++){
                            if(i==j) continue;
                            if(p[i].x-p[i].r<=p[j].x&&p[j].x<=p[i].x+p[i].r)
                                    add(i,j);
                            }*/
    for (int i = ; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    for (int i = ; i <= n; i++) {
        for (int j = first[i]; j; j = edge[j].nex) {
            int y = edge[j].ed;
            if (bl[i] == bl[y])
                continue;
            addc(bl[y], bl[i]);
            du[bl[i]]++;
        }
    }
        /*    for(int i=1;i<=sccnum;i++){
                    for(int j=0;j<scc[i].size();j++)
                            cout<<scc[i][j]<<" ";
                    cout<<endl;
            }*/
    topsort();
    /*    for(int i=1;i<=sccnum;i++)
                    cout<<ans[i]<<" ";cout<<endl;*/
    for (int i = ; i <= n; i++)
        ans1 = (ans1 + 1ll * i * (r[bl[i]]-l[bl[i]]+)) % mod;
    printf("%lld", ans1);
    return ;
}

有点错位，凑合看吧，应该好打。