poj 1364 King[差分约束]

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题目大意：有一个序列。给定一些约束条件，格式为si、ni、oi、ki，意思是序列中第si项到第si+ni项的和>（或<）ki，oi表示>(用"gt”表示)或<(用"lt"表示)。问这样的序列是否存在。存在输出"lamentable kingdomi"，否则输出"successful conspiracy"。

思路：简单的差分约束，就是题目描述有点不太好懂，还有就是，题目给出的是>,<，要把它转化成>=,<=。需要注意，虽然序列长度为n，但是建图的时候用来第0项，相当于序列长度变为了n+1，所以bellman算法要循环n次！

#include<queue>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define M 110#define INF 0x3f3f3f3fstruct node{    int u, v, w;}edge[M];int n, m;int mx, mn;int dist[M];void bellman_ford(){    memset(dist, 0, sizeof(dist));    for (int i = 1; i <= n; i++)//循环n次！    {        for (int i = 0; i <= m; i++)        {            int a = edge[i].u, b = edge[i].v;            if (dist[a] < INF && dist[b] > dist[a] + edge[i].w)            {                dist[b] = dist[a] + edge[i].w;            }        }    }}int main(){    while (cin >> n, n)    {        cin >> m;        //mx = -INF; mn = INF;        for (int i = 1; i <= m; i++)        {            int a, b, k;            string c;            cin >> a >> b >> c >> k;            if (c == "gt")            {                k++;                edge[i].u = a + b;                edge[i].v = a - 1;                edge[i].w = -k;            }            else            {                k--;                edge[i].u = a - 1;                edge[i].v = a + b;                edge[i].w = k;            }        }        bellman_ford();        int flog = 0;        for (int i = 1; i <= m; i++)        {            int a = edge[i].u, b = edge[i].v;            if (dist[a] < INF && dist[b] > dist[a] + edge[i].w)            {                flog = 1;                break;            }        }        if (flog)        {            printf("successful conspiracy\n");        }        else            printf("lamentable kingdom\n");    }    return 0;}

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