afasf,令T[i] = a1+a2+....ai;

则Si = a[Si]+a[Si+1]+...a[Si+ni] = T[Si+ni]-T[Si-1];

以T[i]为节点,i的范围为0<=i<=n,然后根据题目条件加上约束条件,然后判断正/负均可。

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <stack>
using namespace std;

const int maxn = 2010 + 10;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, dist;
    Edge(int from, int to, int dist): from(from), to(to), dist(dist) {}
};

struct SPFA
{
    int n, m;
    
    vector<int> G[maxn];
    vector<Edge> edges;
    
    bool inq[maxn];
    int d[maxn];
    int p[maxn];
    int cnt[maxn];
    
    void init(int n)
    {
        this->n = n;
        for(int i = 0; i <= n; i++) G[i].clear();
        edges.clear();
    }
    
    void AddEdge(int from, int to, int dist)
    {
        edges.push_back(Edge (from, to, dist));
        m = edges.size();
        G[from].push_back(m-1);
    }
    
    bool spfa(int s)
    {
        queue<int> Q;
        memset(inq, 0, sizeof(inq));
        memset(cnt, 0, sizeof(cnt));
        for(int i = 0; i <= n; i++) { inq[s] = 1; Q.push(i); d[i] = 0; }
        while(!Q.empty())
        {
            int u = Q.front(); Q.pop();
            inq[u] = 0;
            for(int i = 0; i < G[u].size(); i++)
            {
                Edge &e = edges[G[u][i]];
                if(d[e.to] < d[u] + e.dist)
                {
                    d[e.to] = d[u]+e.dist;
                    p[e.to] = G[u][i];
                    if(!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to] = 1;
                        if(++cnt[e.to] > n) return 1;
                    }
                }
            }
        }
        return 0;
    }
};

///////////////////
SPFA g;
int n, m;

//////// T[i]的范围是0~n;

int main()
{
    while(~scanf("%d%d", &n, &m) && n)
    {
        g.init(n+10);
        
        int SI, NI, KI; char cmd[9];
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%s%d", &SI, &NI, cmd, &KI);
            if(cmd[0] == 'g') g.AddEdge(SI-1, SI+NI, KI+1);
            else if(cmd[0] == 'l') g.AddEdge(SI+NI, SI-1, -(KI-1));
        }
        if(!g.spfa(0)) printf("lamentable kingdom\n");
        else printf("successful conspiracy\n");
    }
    return 0;
}
View Code

 

 

 

posted on 2013-09-12 18:50  Buck Meister  阅读(326)  评论(0编辑  收藏  举报